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OXIDATION NUMBERS CALCULATOR

To calculate oxidation numbers of elements in the chemical compound, enter it's formula and click 'Calculate' (for example: Ca2+ , HF2^- , Fe4[Fe(CN)6]3 , NH4NO3 , so42- , ch3cooh , cuso4*5h2o ).

The oxidation state of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds. The oxidation number is synonymous with the oxidation state. Determining oxidation numbers from the Lewis structure (Figure 1a) is even easier than deducing it from the molecular formula (Figure 1b). The oxidation number of each atom can be calculated by subtracting the sum of lone pairs and electrons it gains from bonds from the number of valence electrons. Bonds between atoms of the same element (homonuclear bonds) are always divided equally.

When dealing with organic compounds and formulas with multiple atoms of the same element, it's easier to work with molecular formulas and average oxidation numbers (Figure 1d). Organic compounds can be written in such a way that anything that doesn't change before the first C-C bond is replaced with the abbreviation R (Figure 1c). Unlike radicals in organic molecules, R cannot be hydrogen. Since the electrons between two carbon atoms are evenly spread, the R group does not change the oxidation number of the carbon atom it's attached to. You can find examples of usage on the Divide the redox reaction into two half-reactions page.

Rules for assigning oxidation numbers

• The oxidation number of a free element is always 0.
• The oxidation number of a monatomic ion equals the charge of the ion.
• Fluorine in compounds is always assigned an oxidation number of -1.
• The alkali metals (group I) always have an oxidation number of +1.
• The alkaline earth metals (group II) are always assigned an oxidation number of +2.
• Oxygen almost always has an oxidation number of -2, except in peroxides (H 2 O 2 ) where it is -1 and in compounds with fluorine (OF 2 ) where it is +2.
• Hydrogen has an oxidation number of +1 when combined with non-metals, but it has an oxidation number of -1 when combined with metals.
• The algebraic sum of the oxidation numbers of elements in a compound is zero.
• The algebraic sum of the oxidation states in an ion is equal to the charge on the ion.

Assigning oxidation numbers to organic compounds

• cysteine: HO2CCH(NH2)CH2SH

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1.5 – Redox Reactions

Earth’s atmosphere contains about 20% molecular oxygen, O 2 , a chemically reactive gas that plays an essential role in the metabolism of aerobic organisms and in many environmental processes that shape the world. The term oxidation was originally used to describe chemical reactions involving O 2 , a term which you may be familiar with within the context of real-life scenarios and applications like the browning of some fruits and the implication of antioxidants. However, in the sciences, its meaning has evolved to refer to a broad and important reaction class known as oxidation-reduction (redox) reactions. A few examples of such reactions will be used to develop a clear picture of this classification, and we’ll use the stoichiometry skills you’ve learned throughout this chapter to balance redox reactions and solve amounts/concentrations of reactants/products.

Oxidation-Re ductio n (Redox) Reactions

Some redox reactions involve the transfer of electrons between reactant species to yield ionic products, such as the reaction between sodium and chlorine to yield sodium chloride:

2 Na (s) + Cl 2 ( g)   →  2 NaCl (s)

It is helpful to view the process with regard to each individual reactant, that is, to represent the fate of each reactant in the form of an equation called a half-reaction :

2 Na ( s)   →  2 Na + (s) + 2 e –

Cl 2 (g) + 2 e –   →  2 Cl – (s)

These equations show that Na atoms lose electrons while Cl atoms (in the Cl 2 molecule) gain electrons , recall the “ s ”  indicates that the resulting ions are present in the form of a solid ionic compound. For redox reactions of this sort, the loss and gain of electrons define the complementary processes that occur:

Oxidation = loss of electrons

Reduction = gain of electrons

Table 1.5.1. Tips & tricks – Oxidation & reduction memory aids.

In this reaction, then, sodium undergoes oxidation and chlorine undergoes reduction . Viewed from a more active perspective, sodium functions as a reducing agent (reductant ) , since it provides electrons to (or reduces) chlorine. Likewise, chlorine functions as an oxidizing agent (oxidant) , as it effectively removes electrons from (oxidizes) sodium.

Reducing agent = species that is oxidized

Oxidizing agent = species that is reduced

Hence, given that the electrons are transferred from one reactant to another, it’s important to remember that if something has been oxidized (it lost electrons), then something else has been reduced (gained those electrons).

Some redox processes, however, do not involve the transfer of electrons. Consider, for example, a reaction similar to the one yielding HCl:

H 2 (g) + Cl 2 ( g)   →  2 HCl (g)

The product of this reaction is a covalent compound, so the transfer of electrons in the explicit sense is not involved. To clarify the similarity of this reaction to the previous one and permit an unambiguous definition of redox reactions, a property called oxidation number has been defined. The oxidation number (or oxidation state ) of an element in a compound is the charge its atoms would possess if the compound was ionic . The following guidelines are used to assign oxidation numbers to each element in a molecule or ion:

i) The oxidation number of an atom in its elemental form is zero (e.g. O 2 , Cl 2 , Na).

ii) The oxidation number of a monatomic ion is equal to the ion’s charge (e.g. +1 for Na + , -2 for O 2- ).

iii) The sum of oxidation numbers for all atoms in a molecule or polyatomic ion equals the charge on the molecule or ion.

Oxidation numbers for common nonmetals are usually assigned as follows:

A) Hydrogen: +1 when combined with nonmetals (e.g. H 2 O), −1 when combined with metals and boron

B) Oxygen: −2 in most compounds (e.g. H 2 O), sometimes −1 (so-called peroxides, O 2 2− ), very rarely −1/2 (so-called superoxides, O 2 − ), might be +2 or -1 when coupled to a more electronegative centre (such as F) or a group 1 or group 2 metal

C) Halogens: −1 for F always (e.g. HF), −1 for other halogens when combined with metals, nonmetals (except O), and other halogens lower in the group

Note: The proper convention for reporting charge is to write the number first, followed by the sign (e.g., 2+), while oxidation number is written with the reversed sequence, sign followed by number (e.g., +2). This convention aims to emphasize the distinction between these two related properties.

A few tips to keep in mind as you solve problems involving determining oxidation numbers in compounds: 1) if two rules appear to contradict each other, follow the rule that appears higher on the list; 2) for a multi-atom species, figure out the easy oxidation states first, then solve for the other unknown atoms.

Example 1.5.1 – Assigning Oxidation Numbers

Follow the guidelines in this section of the text to assign oxidation numbers to all the elements in the following species:

(b) SO 3 2-

(c) Na 2 SO 4

(a) According to guideline A above, the oxidation number for H is +1.

Using this oxidation number and the compound’s formula, guideline iii may then be used to calculate the oxidation number for sulfur, represented by x:

Charge on H 2 S = 0 = (2 + 1) + (1 × x )

x = 0 – (2 × (+1)) = -2

(b) Guideline B suggests the oxidation number for oxygen is −2.

Using this oxidation number and the ion’s formula, guideline iii may then be used to calculate the oxidation number for sulfur:

Charge on SO 3 2- =   -2 = (3 × (-2)) + (1 × x )

x = -2 – (3 × (-2)) = +4

(c) For ionic compounds, it’s convenient to assign oxidation numbers for the cation and anion separately.

According to guideline ii, the oxidation number for sodium is +1.

Assuming the usual oxidation number for oxygen (−2 per guideline B), the oxidation number for sulfur is calculated as directed by guideline iii:

charge on SO 4 2-  = -2 = (4 × (-2)) + (1 × x )

x = -2 – (4 × (-2)) = +6

Check Your Learning 1.5.1 – Assigning Oxidation Numbers

Assign oxidation states to the elements whose atoms are underlined in each of the following compounds or ions:

(a) K N O 3

(c) N H 4 +

(d) H 2 P O 4 −

(a) N, +5; (b) Al, +3; (c) N, −3; (d) P, +5

Using the oxidation number concept, an all-inclusive definition of redox reaction has been established. Oxidation-reduction (redox) reactions are those in which one or more elements involved undergo a change in oxidation number. (While the vast majority of redox reactions involve changes in oxidation number for two or more elements, a few interesting exceptions to this rule do exist – see Example 1.5.2 “Describing Redox Reactions”.) Definitions for the complementary processes of this reaction class are correspondingly revised as shown here:

Oxidation = increase in oxidation number

Reduction = decrease in oxidation number

Returning to the reactions used to introduce this topic, they may now both be identified as redox processes. In the reaction between sodium and chlorine to yield sodium chloride, sodium is oxidized (its oxidation number increases from 0 in Na to +1 in NaCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl 2 to −1 in NaCl). In the reaction between molecular hydrogen and chlorine, hydrogen is oxidized (its oxidation number increases from 0 in H 2 to +1 in HCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl 2 to −1 in HCl).

Types of Redox Reactions – Combustion

Several subclasses of redox reactions are recognized, including combustion reactions in which the reductant (also called a fuel ) and oxidant (often, but not necessarily, molecular oxygen) react vigorously and produce significant amounts of heat, and often light, in the form of a flame. Solid rocket-fuel reactions (Figure 1.5.1) are combustion processes. A typical propellant reaction in which solid aluminum is oxidized by ammonium perchlorate is represented by this equation:

10 Al (s) + 6 NH 4 ClO 4 ( s)   →  4 Al 2 O 3 (s) + 2 AlCl 3 (s) + 12 H 2 O (g) + 3 N 2 (g)

Figure 1.5.1. Many modern rocket fuels are solid mixtures of substances combined in carefully measured amounts and ignited to yield a thrust-generating chemical reaction. (credit: modification of work by NASA)

Types of Redox Reactions – Single-displacement (Replacement)

Single-displacement (replacement) reactions are redox reactions in which an ion in solution is displaced (or replaced) via the oxidation of a metallic element. One common example of this type of reaction is the acid oxidation of certain metals:

Zn (s) + 2 HCl ( aq )   →  ZnCl 2 ( aq ) + H 2 (g)

Metallic elements may also be oxidized by solutions of other metal salts; for example:

Cu (s) + 2 AgNO 3 ( aq )   →  Cu(NO 3 ) 2 ( aq ) + 2 Ag (s)

This reaction may be observed by placing copper wire in a solution containing a dissolved silver salt. Silver ions in solution are reduced to elemental silver at the surface of the copper wire, and the resulting Cu 2+ ions dissolve in the solution to yield a characteristic blue colour (Figure 1.5.2).

Figure 1.5.2. (a) A copper wire is shown next to a solution containing silver(I) ions. (b) Displacement of dissolved silver ions by copper ions results in (c) accumulation of gray-coloured silver metal on the wire and development of a blue colour in the solution, due to dissolved copper ions. (credit: modification of work by Mark Ott)

Example 1.5.2 – Describing Redox Reactions

Identify which equations represent redox reactions, providing a name for the reaction if appropriate. For those reactions identified as redox, name the oxidant and reductant.

(a) ZnCO 3 ( s)   →  ZnO (s) + CO 2 (g)

(b) 2 Ga (l) + 3 Br 2 (l )   →  2 GaBr 3 (s)

(c) 2 H 2 O 2 ( aq )   →  2 H 2 O (l) + O 2 (g)

(d) BaCl 2 ( aq ) + K 2 SO 4 ( aq )   →  BaSO 4 (s) + 2 KCl ( aq )

(e) C 2 H 4 (g) + 3 O 2 ( g)   →  2 CO 2 (g) + 2 H 2 O (l) 

Redox reactions are identified per definition if one or more elements undergo a change in oxidation number.

(a) This is not a redox reaction, since oxidation numbers remain unchanged for all elements.

(b) This is a redox reaction. Gallium is oxidized, its oxidation number increasing from 0 in Ga ( l ) to +3 in GaBr 3 ( s ). The reducing agent is Ga ( l ). Bromine is reduced, its oxidation number decreasing from 0 in Br 2 ( l ) to −1 in GaBr 3 ( s ). The oxidizing agent is Br 2 ( l ).

(c) This is a redox reaction. It is a particularly interesting process, as it involves the same element, oxygen, undergoing both oxidation and reduction (a so-called disproportionation reaction) . Oxygen is oxidized, its oxidation number increasing from −1 in H 2 O 2 ( aq ) to 0 in O 2 ( g ). Oxygen is also reduced, its oxidation number decreasing from −1 in H 2 O 2 ( aq ) to −2 in H 2 O ( l ). For disproportionation reactions, the same substance functions as an oxidant and a reductant.

(d) This is not a redox reaction since oxidation numbers remain unchanged for all elements.

(e) This is a redox reaction (combustion). Carbon is oxidized, its oxidation number increasing from −2 in C 2 H 4 ( g ) to +4 in CO 2 ( g ). The reducing agent (fuel) is C 2 H 4 ( g ). Oxygen is reduced, its oxidation number decreasing from 0 in O 2 ( g ) to −2 in H 2 O ( l ). The oxidizing agent is O 2 ( g ).

Check Your Learning 1.5.2 – Describing Redox Reactions

This equation describes the production of tin (II) chloride:

Sn (s) + 2 HCl ( g)   →  SnCl 2 (s) + H 2 (g)

Is this a redox reaction? If so, provide a more specific name for the reaction if appropriate, and identify the oxidant and reductant.

Yes, a single-replacement reaction. Sn ( s ) is the reductant, HCl ( g ) is the

Balancing Redox Reactions – The Half-Reaction Method

The redox reactions discussed so far tended to be rather simple, and conservation of mass (atom counting by type) and deriving a correctly balanced chemical equation was relatively simple. Balancing other reactions becomes more complicated when they take place in aqueous media that often involves water, hydronium ions, and hydroxide ions as reactants or products. Although these species are not oxidized or reduced, they do participate in chemical change in other ways (e.g., by providing the elements required to form oxyanions). Equations representing these reactions are sometimes very difficult to balance by inspection, so systematic approaches have been developed to assist in the process. One very useful approach is to use the half-reaction method , which splits oxidation-reduction reactions into their oxidation “half” and reduction “half” to make finding the overall equation easier. This method involves the following general steps:

1. Write out the net ionic form of the reaction (see the preceding section “Solution Stoichiometry” to refresh your memory on expressing net ionic equations).

2. Separate this equation into its two half-reactions representing the redox process.

3. Balance all elements except oxygen and hydrogen.

4. Balance oxygen atoms by adding H 2 O molecules.

5. Balance hydrogen atoms by adding H + ions.

6. Balance the net charge on each side by adding electrons.

7. If necessary, multiply each half-reaction’s coefficients by the smallest possible integers to yield equal numbers of electrons in each.

8. Add the balanced half-reactions together and simplify by removing species that appear on both sides of the equation.

9. Verify your final reaction (check that the number of atoms and the total charges are balanced).

Redox reactions frequently occur in solutions, which could be acidic, basic, or neutral. When balancing oxidation-reduction reactions, the nature of the solution may be important:

Acidic solution: follow all steps 1-9 as normal.

Basic solution: follow all steps 1-9 as normal, add two additional “steps” to step 5 :

• Add OH − ions to both sides of the equation in numbers equal to the number of H + ions. A common mistake students make is that they add OH – to only one side of the equation, but just like in math equations, if you add one element to one side, you must add it to the other side as well (hence, add the same number of OH – ions to both sides!).
• On the side of the equation containing both H + and OH − ions, combine these ions to yield water molecules.

Example 1.5.3 – Balancing Redox Reactions in Acidic Solution 

Write a balanced equation for the reaction between dichromate ion and iron (II) to yield iron(III) and chromium(III) in an acidic solution.

Cr 2 O 7 2- + Fe 2 +   →  Cr 3+ + Fe 3+

1. Write the two half-reactions . Each half-reaction will contain one reactant

and one product with one element in common.

Fe 2 +   →  Fe 3+

Cr 2 O 7 2 –  →  Cr 3+

2. Balance all elements except oxygen and hydrogen . The iron half-reaction is already balanced, but the chromium half-reaction shows two Cr atoms on the left and one Cr atom on the right. Changing the coefficient on the right side of the equation to 2 achieves balance with regard to Cr atoms.

Fe 2 +  →  Fe 3+

Cr 2 O 7 2 –   →  2 Cr 3+

3. Balance oxygen atoms by adding H 2 O molecules . The iron half-reaction does not contain O atoms. The chromium half-reaction shows seven O atoms on the left and none on the right, so seven water molecules are added to the right side.

Cr 2 O 7 2 –  →  2 Cr 3+ + 7 H 2 O

4. Balance hydrogen atoms by adding H + ions.  The iron half-reaction does not contain H atoms. The chromium half-reaction shows 14 H atoms on the right and none on the left, so 14 hydrogen ions are added to the left side.

Cr 2 O 7 2- + 14 H +   →  2 Cr 3+ + 7 H 2 O

5. Balance charge by adding electrons . The iron half-reaction shows a total charge of 2+ on the left side (1 Fe 2+ ion) and 3+ on the right side (1 Fe 3+ ion). Adding one electron to the right side brings that side’s total charge to (3+) + (1−) = 2+, and charge balance is achieved. The chromium half-reaction shows a total charge of (1 × 2−) + (14 × 1+) = 12+ on the left side (1 Cr 2 O 7 2− ion and 14 H + ions). The total charge on the right side is (2 × 3+) = 6 + (2 Cr 3+ ions). Adding six electrons to the left side will bring that side’s total charge to ((12+) + (6−)) = 6+, and charge balance is achieved.

Fe 2 +   →  Fe 3+ + e –

Cr 2 O 7 2- + 14 H + + 6 e –   →  2 Cr 3+ + 7 H 2 O

Note : At this point, make sure you always check that one of your equations has the electron(s) on the reactants side, while the other equation has the electron(s) on the products side. If this isn’t the case, it’s a clear sign that you’ve done something wrong – go back and check your work to find the error(s).

6. Multiply the two half-reactions so the number of electrons in one reaction equals the number of electrons in the other reaction . To be consistent with mass conservation, and the idea that redox reactions involve the transfer (not creation or destruction) of electrons, the iron half-reaction’s coefficient must be multiplied by 6.

6 Fe 2 +  →  6 Fe 3+ + 6 e –

A good way to qualitatively check your work when balancing both half-reactions is by recalling your understanding of oxidation and reduction. For the first reaction, 6 Fe 2+ → 6 Fe 3+ + 6 e – , the oxidation state of iron goes up from +2 and +3, and there are electrons being lost in the products side, so it’s the oxidation half-reaction. For the second reaction, the oxidation state of chromium goes down from +6 to +3, and electrons are gained in the reactants side, so it’s the reduction half-reaction.

7. Add the balanced half-reactions and cancel species that appear on both sides of the equation .

6 Fe 2+ + Cr 2 O 7 2- + 14 H + + 6 e –   →  6 Fe 3+ + 6 e – + 2 Cr 3+ + 7 H 2 O

Only the six electrons are redundant species. Removing them from each side of the equation yields the simplified, balanced equation here:

6 Fe 2+ + Cr 2 O 7 2- + 14 H +   →  6 Fe 3+ + 2 Cr 3+ + 7 H 2 O

A final check of atom and charge balance confirms the equation is balanced.

Check Your Learning 1.5. 3 – Balancing Redox Reactions in Acidic Solution 

In acidic solution, hydrogen peroxide reacts with Fe 2+ to produce Fe 3+ and H 2 O. Write a balanced equation for this reaction.

H 2 O 2 ( aq ) + 2 H + ( aq ) + 2 Fe 2 +  →  2 H 2 O (l) + 2 Fe 3+

Example 1.5. 4 – Balancing Redox Reactions in Basic Solution

Balance the following reaction equation in basic solution:

MnO 4 – ( aq ) + Cr(OH) 3 (s)   →  MnO 2 (s) + CrO 4 2- ( aq )

This is an oxidation-reduction reaction, so start by collecting the species given into an unbalanced oxidation half-reaction and an unbalanced reduction half-reaction:

Oxidation (unbalanced): Cr(OH) 3 (s)   →  CrO 4 2- ( aq )

Reduction (unbalanced): MnO 4 – ( aq )   →  MnO 2 (s)

Starting with the oxidation half-reaction, we can balance the chromium:

Oxidation (unbalanced): Cr(OH) 3 (s)  →  CrO 4 2- ( aq )

In acidic solution, we can use or generate hydrogen ions (H + ). Adding one water molecule to the left side provides the necessary oxygen; the “leftover” hydrogen appears as five H + on the right side:

Oxidation (unbalanced): Cr(OH) 3 (s) + H 2 O (l)  →  CrO 4 2- ( aq ) + 5 H + ( aq )

The left side of the equation has a total charge of [0], and the right side a total charge of [−2 + 5 × (+1) = +3]. The difference is three, adding three electrons to the right side produces a mass- and charge-balanced oxidation half-reaction (in acidic solution):

Oxidation (balanced): Cr(OH) 3 (s) + H 2 O (l)   →  CrO 4 2- ( aq ) + 5 H + ( aq ) + 3 e –

Checking the half-reaction:

  Cr: Does (1 x 1) = (1 x 1)? Yes.

  H: Does (1 x 3 + 1 x 2) = (5 x 1)? Yes.

  O: Does (1 x 3 + 1 x 1) = (4 x 1)? Yes.

  Charge: Does [0] = [1 x (-2) +5 x (+1) + 3 x (-1)]? Yes.

Now work on the reduction. It is necessary to convert the four O atoms in the MnO 4 − minus the two O atoms in MnO 2 into two water molecules. To do this, add four H + to convert the oxygen into two water molecules:

Reduction (unbalanced): MnO 4 – ( aq ) + 4 H + ( aq )   →  MnO 2 (s) + 2 H 2 O (l)

Then add three electrons to the left side to balance the charge:

Reduction (balanced): MnO 4 – ( aq ) + 4 H + ( aq ) + 3 e –   →  MnO 2 (s) + 2 H 2 O (l)

Make sure to check the half-reaction:

Mn: Does (1 x 1) = (1 x 1)? Yes.

H: Does (4 x 1) = (2 x 2)? Yes.

O: Does (1 x 4) = (1 x 2 + 2 x 1)? Yes.

Charge: Does [1 x (-1) + 4 x (+1) + 3 x (-1)] = [0]? Yes.

Collecting what we have so far:

In this case, both half-reactions involve the same number of electrons; therefore, simply add the two half-reactions together.

Cr(OH) 3 (s) + H 2 O (l) + MnO 4 – ( aq ) + 4 H + ( aq ) + 3 e –   →  MnO 2 (s) + 2 H 2 O (l) + CrO 4 2- ( aq ) + 5 H + ( aq ) + 3 e –

Cr(OH) 3 (s) + MnO 4 – ( aq )   →  MnO 2 (s) + H 2 O (l) + CrO 4 2- ( aq ) + H + ( aq )

Checking each side of the equation:

Cr: Does (1 x 1) = (1 x 1)? Yes.

H: Does (1 x 3) = (2 x 1 + 1 x 1)? Yes.

O: Does (1 x 4 + 1 x 3) = (1 x 4 + 1 x 2 + 1 x 1)? Yes.

Charge: Does [1 x (-1)] = [1 x (-2) + 1 x (+1)]? Yes.

This is the balanced equation in acidic solution. For a basic solution, add one hydroxide ion to each side and simplify:

OH – ( aq ) + Cr(OH) 3 (s) + H 2 O (l) + MnO 4 – ( aq )   →  MnO 2 (s) + H 2 O (l) + CrO 4 2- ( aq ) + (H + + OH – ) ( aq )

OH – ( aq ) + Cr(OH) 3 (s) + H 2 O (l) + MnO 4 – ( aq )   →  MnO 2 (s) + 2H 2 O (l) + CrO 4 2- ( aq )

H: Does (1 x 1 + 1 x 3) = (2 x 2)? Yes.

O: Does (1 x 1 +1 x 4 + 1 x 3) = (1 x 4 + 1 x 2 + 2 x 1)? Yes.

Charge: Does [1 x (-1) + 1 x (-1)] = [1 x (-2)]? Yes.

This is the balanced equation in basic solution.

Check Your Learning 1.5. 4 – Balancing Redox Reactions in Basic Solution

Balance the following in the type of solution indicated.

(a) H 2 + Cu +   ⟶  Cu (acidic solution)

(b) H 2 + Cu(OH) 2   ⟶  Cu (basic solution)

(c) Fe + Ag +  ⟶  Fe 2+ + Ag

(d) Identify the oxidizing agents in reactions (a), (b), and (c).

(e) Identify the reducing agents in reactions (a), (b), and (c).

(a) H 2 ( g ) + Cu 2+ ( aq )  ⟶  2 H + ( aq ) + Cu ( s );

(b) H 2 ( g ) + Cu(OH) 2 ( s )  ⟶  2 H 2 O ( l ) + Cu ( s );

(c) Fe ( s ) + 2 Ag + ( aq )  ⟶  Fe 2+ ( aq ) + 2 Ag ( s );

(d) Recall oxidizing agent = species reduced: (a) Cu 2+ ; (b) Cu(OH) 2 ; (c) Ag +

(e) Recall reducing agent = species oxidized: (a) H 2 ; (b) H 2 ; (c) Fe

★ Questions

1. Determine the oxidation states of the elements in the following compounds:

(e) Mg 2 Si

(f) RbO 2 , rubidium superoxide

2. Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.

(a)H 3 PO 4

(b) Al(OH) 3

(e) In 2 S 3

(f.) P 4 O 6

3. Identify the atoms that are oxidized and reduced, the change in oxidation state for each, and the oxidizing and reducing agents in each of the following equations:

(a) Mg (s) + NiCl 2 ( aq )   →  MgCl 2 ( aq ) + Ni (s)

(b) PCl 3 (l) + Cl 2 ( g)   →  PCl 5 (s)

(c) C 2 H 4 (g) + 3 O 2 ( g)   →  2 CO 2 (g) + 2 H 2 O (g)

(d) Zn (s) + H 2 SO 4 ( aq )   →  ZnSO 4 ( aq ) + H 2 (g)

(e) 2 K 2 S 2 O 3 (s) + I 2 ( s)   →  K 2 S 4 O 6 (s) + 2 KI (s)

(f) 3 Cu (s) + 8 HNO 3 ( aq )   →  3 Cu(NO 3 ) 2 ( aq ) + 2 NO (g) + 4 H 2 O (l)

4. Complete and balance the following oxidation-reduction reactions, which give the highest possible oxidation state for the oxidized atoms.

(a) Al (s) + F 2 ( g)   →

(b) Al (s) + CuBr 2 ( aq )   → (single displacement)

(c) P 4 (s) + O 2 ( g)   →

(d) Ca (s) + H 2 O (l )   →  (products are a strong base and a diatomic gas)

5. When heated to 700–800 °C, diamonds, which are pure carbon, are oxidized by atmospheric oxygen. (They burn!) Write the balanced equation for this reaction.

6. The military has experimented with lasers that produce very intense light when fluorine combines explosively with hydrogen. What is the balanced equation for this reaction?

★★ Questions

7. Complete and balance each of the following half-reactions (steps 2–5 in half-reaction method):

(a) Sn 4+ ( aq )  →  Sn 2+ ( aq )

(b) [Ag(NH 3 ) 2 ] ( aq )   →  Ag + (s) + NH 3 ( aq )

(c) Hg 2 Cl 2 ( s)   →  Hg (l) + Cl – ( aq )

(d) H 2 O (l )   →  O 2 (g) (in acidic solution)

(e) IO 3 – ( aq )   →  I 2 (s)

(f) SO 3 2- ( aq )   →  SO 4 2- ( aq ) (in acidic solution)

(g) MnO 4 – ( aq )   →  Mn 2+ ( aq ) (in acidic solution)

(h) Cl – ( aq )  →  ClO 3 – ( aq ) (in basic solution)

8. Balance each of the following equations according to the half-reaction method:

(a) Sn 2+ ( aq ) + Cu 2+ ( aq )   →   Sn 4+ ( aq ) + Cu + ( aq )

(b) H 2 S (g) + Hg 2 2+ ( aq )   →  Hg (l) + S (s)   (in acid)

(c) CN – ( aq ) + ClO 2 ( aq )   →  CNO – ( aq ) + Cl – ( aq )   (in acid)

(d) Fe 2+ ( aq ) + Ce 4+ ( aq )   →   Fe 3+ ( aq ) + Ce 3+ ( aq )

(e) HBrO ( aq )   →  Br – ( aq ) + O 2 ( g )  (in acid)

9. Balance each of the following equations according to the half-reaction method:

(a) MnO 4 – ( aq ) + NO 2 – ( aq )   →  MnO 2 (s) + NO 3 – ( aq )   (in base)

(b) MnO 4 2- ( aq )  →  MnO 4 – ( aq ) + MnO 2 (s)   (in base)

(c) Br 2 (l) + SO 2 ( g)   →  Br – ( aq ) + SO 4 2- ( aq )   (in acid)

10. Balance the following in acidic solution:

(a) H 2 O 2 + Sn 2 +   →  H 2 O + Sn 4+

(b) PbO 2 + Hg  →  Hg 2 2+ + Pb 2+

(c) Al + Cr 2 O 7 2 –   →  Al 3 + + Cr 3+

11. Balance the following in basic solution:

(a) SO 3 2- ( aq ) + Cu(OH) 2 (s)   →  SO 4 2- ( aq ) + Cu(OH)

(b) O 2 (g) + Mn(OH) 2 (s)   →  MnO 2 (s)

(c) NO 3 – ( aq ) + H 2 ( g)   →  NO (g)

(d) Al (s) + CrO 4 2- ( aq )   →  Al(OH) 3 (s) + Cr(OH) 4 – ( aq )

1. (a) Na +1, I -1; (b) Gd +3, Cl -1; (c) Li +1, N +5, O -3; (d) H +1, Se -2; (e) Mg +2, Si -4; (f) Rb -1, O +0.5; (g) H +1, F -1

2. (a) H +1, P +5, O −2; (b) Al +3, H +1, O −2; (c) Se +4, O −2; (d) K +1, N +3, O −2; (e) In +3, S −2; (f) P +3, O −2

3. (a) Mg: oxidized, 0 → +2, reducing agent, Ni: reduced, +2 → 0, oxidizing agent;

(b) P: oxidized, +3 → 5, reducing agent, Cl:  reduced, 0 → -1, oxidizing agent;

(c) C: oxidized, -2 → +4, reducing agent, O: reduced, 0 → -2, oxidizing agent;

(d) Zn: oxidized, 0 → +2, reducing agent, H: reduced, +1 → 0, oxidizing agent;

(e) S: oxidized, +2 → +5/2, reducing agent, I: reduced, 0 → -1, oxidizing agent;

(f) Cu: oxidized, 0 → +2, reducing agent, N: reduced, +5 → +2, oxidizing agent

4. (a) 2 Al (s) + 3 F 2 ( g)  →  2 AlF 3 (s)

(b) Al (s) + CuBr 2 ( aq )  →  3 Cu (s) + 2 AlBr 3 ( aq )

(c) P 4 (s) + O 2 ( g)   →  P 4 O 10 (s)

(d) Ca (s) + H 2 O (l )   →  Ca(OH) 2 ( aq ) + H 2 (g)

5. C diamond (s) + O 2 ( g)   →  CO 2 (g)

6. H 2 (g) + F 2 ( g)   →  CO 2 (g)

7. (a) Sn 4+ ( aq ) + 2e –   →  Sn 2+ ( aq )

(b) [Ag(NH 3 ) 2 ] ( aq ) + e –   →  Ag + (s) + NH 3 ( aq )

(c) Hg 2 Cl 2 (s) + 2 e –   →  Hg (l) + Cl – ( aq )

(d) 2 H 2 O (l )   →  O 2 (g) + 4 H + ( aq ) + 4 e –

(e) 6 H 2 O (l) + IO 3 – ( aq ) + 10 e –  →  I 2 (s) + 12 OH – ( aq )

(f) H 2 O (l) + SO 3 2- ( aq )   →  SO 4 2- ( aq ) + 2 H + ( aq ) + 2 e –

(g) 8 H + ( aq ) + MnO 4 – ( aq ) + 5 e –  →  Mn 2+ ( aq ) + 4 H 2 O (l)

(h) Cl – ( aq ) + 6 OH – ( aq )  →  ClO 3 – ( aq ) + 3 H 2 O (l) + 6 e –

8. (a) Sn 2+ ( aq ) + 2 Cu 2+ ( aq )   →  Sn 4+ ( aq ) + 2 Cu + ( aq )

(b) H 2 S (g) + Hg 2 2+ ( aq ) + 2 H 2 O (l )   →  2 Hg (l) + S (s)  + 2 H 3 O + ( aq )

(c) 5 CN – ( aq ) + 2 ClO 2 ( aq ) + 3 H 2 O (l ) → 5 CNO – ( aq ) + 2 Cl – ( aq )  + 2 H 3 O + ( aq )

(d) Fe 2+ ( aq ) + Ce 4+ ( aq )   →  Fe 3+ ( aq ) + Ce 3+ ( aq )

(e) 2   HBrO ( aq ) + 2 H 2 O (l )  →  2 H 3 O + ( aq ) + 2 Br – ( aq ) + O 2 ( g ) 

9. (a) 2 MnO 4 – ( aq ) + 3 NO 2 – ( aq ) + H 2 O (l )  →  2 MnO 2 (s) + 3 NO 3 – ( aq )  + 2 OH – ( aq )

(b) 3 MnO 4 2- ( aq ) + 2 H 2 O (l)   →  2 MnO 4 – ( aq ) + MnO 2 (s) + 4 OH – ( aq )

(c) Br 2 ( l ) + SO 2 (g) + 2 H 2 O (l )   →  2 Br – ( aq ) + SO 4 2- ( aq )  + 4 H + ( aq )

10. (a) 2 H + + H 2 O 2 + Sn 2 +   →  H 2 O + Sn 4+

(b) 4 H + + PbO 2 + 2 Hg  →  Hg 2 2+ + Pb 2+ + 2 H 2 O

(c) 2 Al + Cr 2 O 7 2- + 14 H +   →  2 Al 3+ + 2 Cr 3+ + 7 H 2 O

11. (a) SO 3 2- ( aq ) + 2 Cu(OH) 2 (s)   →  SO 4 2- ( aq ) + 2 Cu(OH) + H 2 O

(b) O 2 (g) + 2 Mn(OH) 2 (s)   →  2 MnO 2 (s) + 2 H 2 O

(c) 2 NO 3 – ( aq ) + 3 H 2 (g) + 2 H 2 O  →  NO (g) + 2 OH – ( aq )

(d) Al (s) + CrO 4 2- ( aq ) + 4 H 2 O  →  Al(OH) 3 (s) + Cr(OH) 4 – ( aq ) + OH – ( aq )

Loss of one or more electrons by an atom; an increase in oxidation number

The individual oxidation or reduction reaction of a redox reaction

The gain of one or more electrons by an atom; a decrease in oxidation number

A substance that brings about the reduction of another substance, and in the process becomes oxidized

A substance that brings about the oxidation of another substance, and in the process becomes reduced

The charge each atom of an element would have in a compound if the compound were ionic

A chemical reaction that involves the transfer of electrons, hence it involves a change in oxidation number for one or more reactant elements

Redox chemical reaction in which a reactant combines with oxygen to produce oxides of all other elements as products; produces significant amounts of energy in the form of heat and, sometimes, light

Redox reaction involving the oxidation of an elemental substance by an ionic species; one element is substituted for another element in a compound

Method of balancing redox reactions by writing and balancing the individual half-reactions

General Chemistry for Gee-Gees Copyright © by Kevin Roy; Mahdi Zeghal; Jessica M. Thomas; and Kathy-Sarah Focsaneanu is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Oxidation States (Oxidation Numbers)

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• Truro School in Cornwall

Oxidation states simplify the process of determining what is being oxidized and what is being reduced in redox reactions. However, for the purposes of this introduction, it would be useful to review and be familiar with the following concepts:

• oxidation and reduction in terms of electron transfer
• electron-half-equations

To illustrate this concept, consider the element vanadium, which forms a number of different ions (e.g., \(\ce{V^{2+}}\) and \(\ce{V^{3+}}\)). The 2+ ion will be formed from vanadium metal by oxidizing the metal and removing two electrons:

\[ \ce{V \rightarrow V^{2+} + 2e^{-}} \label{1}\]

The vanadium in the \( \ce{V^{2+}}\) ion has an oxidation state of +2. Removal of another electron gives the \(\ce{V^{3+}}\) ion:

\[ \ce{V^{2+} \rightarrow V^{3+} + e^{-}} \label{2}\]

The vanadium in the \(\ce{V^{3+} }\) ion has an oxidation state of +3. Removal of another electron forms the ion \(\ce{VO2+}\):

\[ \ce{V^{3+} + H_2O \rightarrow VO^{2+} + 2H^{+} + e^{-}} \label{3}\]

The vanadium in the \(\ce{VO^{2+}}\) is now in an oxidation state of +4.

Notice that the oxidation state is not always the same as the charge on the ion (true for the products in Equations \ref{1} and \ref{2}), but not for the ion in Equation \ref{3}).

The positive oxidation state is the total number of electrons removed from the elemental state. It is possible to remove a fifth electron to form another the \(\ce{VO_2^{+}}\) ion with the vanadium in a +5 oxidation state.

\[ \ce{VO^{2+} + H_2O \rightarrow VO_2^{+} + 2H^{+} + e^{-}}\]

Each time the vanadium is oxidized (and loses another electron), its oxidation state increases by 1. If the process is reversed, or electrons are added, the oxidation state decreases. The ion could be reduced back to elemental vanadium, with an oxidation state of zero.

If electrons are added to an elemental species, its oxidation number becomes negative. This is impossible for vanadium, but is common for nonmetals such as sulfur:

\[ \ce{S + 2e^- \rightarrow S^{2-}} \]

Here the sulfur has an oxidation state of -2.

The oxidation state of an atom is equal to the total number of electrons which have been removed from an element (producing a positive oxidation state) or added to an element (producing a negative oxidation state) to reach its present state.

• Oxidation involves an increase in oxidation state
• Reduction involves a decrease in oxidation state

Recognizing this simple pattern is the key to understanding the concept of oxidation states. The change in oxidation state of an element during a reaction determines whether it has been oxidized or reduced without the use of electron-half-equations.

Determining oxidation states

Counting the number of electrons transferred is an inefficient and time-consuming way of determining oxidation states. These rules provide a simpler method.

Rules to determine oxidation states

• The oxidation state of an uncombined element is zero. This applies regardless of the structure of the element: Xe, Cl 2 , S 8 , and large structures of carbon or silicon each have an oxidation state of zero.
• The sum of the oxidation states of all the atoms or ions in a neutral compound is zero.
• The sum of the oxidation states of all the atoms in an ion is equal to the charge on the ion.
• The more electronegative element in a substance is assigned a negative oxidation state. The less electronegative element is assigned a positive oxidation state. Remember that electronegativity is greatest at the top-right of the periodic table and decreases toward the bottom-left.
• Some elements almost always have the same oxidation states in their compounds:

The reasons for the exceptions

Hydrogen in the metal hydrides : Metal hydrides include compounds like sodium hydride, NaH. Here the hydrogen exists as a hydride ion, H - . The oxidation state of a simple ion like hydride is equal to the charge on the ion—in this case, -1.

Alternatively, the sum of the oxidation states in a neutral compound is zero. Because Group 1 metals always have an oxidation state of +1 in their compounds, it follows that the hydrogen must have an oxidation state of -1 (+1 -1 = 0).

Oxygen in peroxides : Peroxides include hydrogen peroxide, H 2 O 2 . This is an electrically neutral compound, so the sum of the oxidation states of the hydrogen and oxygen must be zero.

Because each hydrogen has an oxidation state of +1, each oxygen must have an oxidation state of -1 to balance it.

Oxygen in F 2 O : The deviation here stems from the fact that oxygen is less electronegative than fluorine; the fluorine takes priority with an oxidation state of -1. Because the compound is neutral, the oxygen has an oxidation state of +2.

Chlorine in compounds with fluorine or oxygen : Because chlorine adopts such a wide variety of oxidation states in these compounds, it is safer to simply remember that its oxidation state is not -1, and work the correct state out using fluorine or oxygen as a reference. An example of this situation is given below.

Example \(\PageIndex{1}\): Chromium

What is the oxidation state of chromium in Cr 2 + ?

For a simple ion such as this, the oxidation state equals the charge on the ion: +2 (by convention, the + sign is always included to avoid confusion)

What is the oxidation state of chromium in CrCl 3 ?

This is a neutral compound, so the sum of the oxidation states is zero. Chlorine has an oxidation state of -1 (no fluorine or oxygen atoms are present). Let n equal the oxidation state of chromium:

n + 3(-1) = 0

The oxidation state of chromium is +3.

Example \(\PageIndex{2}\): Chromium

What is the oxidation state of chromium in Cr(H 2 O) 6 3+ ?

This is an ion and so the sum of the oxidation states is equal to the charge on the ion. There is a short-cut for working out oxidation states in complex ions like this where the metal atom is surrounded by electrically neutral molecules like water or ammonia.

The sum of the oxidation states in the attached neutral molecule must be zero. That means that you can ignore them when you do the sum. This would be essentially the same as an unattached chromium ion, Cr 3 + . The oxidation state is +3.

What is the oxidation state of chromium in the dichromate ion, Cr 2 O 7 2 - ?

The oxidation state of the oxygen is -2, and the sum of the oxidation states is equal to the charge on the ion. Don't forget that there are 2 chromium atoms present.

2n + 7(-2) = -2

Example \(\PageIndex{3}\): Copper

What is the oxidation state of copper in CuSO 4 ?

Unfortunately, it isn't always possible to work out oxidation states by a simple use of the rules above. The problem in this case is that the compound contains two elements (the copper and the sulfur) with variable oxidation states.

In cases like these, some chemical intuition is useful. Here are two ways of approaching this problem:

• Recognize CuSO 4 as an ionic compound containing a copper ion and a sulfate ion, SO 4 2 - . To form an electrically neutral compound, the copper must be present as a Cu 2+ ion. The oxidation state is therefore +2.
• Recognize the formula as being copper(II) sulfate (the (II) designation indicates that copper is in a +2 oxidation state, as discussed below).

Using oxidation states

In naming compounds.

You will have come across names like iron(II) sulfate and iron(III) chloride. The (II) and (III) are the oxidation states of the iron in the two compounds: +2 and +3 respectively. That tells you that they contain Fe 2 + and Fe 3 + ions.

This can also be extended to negative ions. Iron(II) sulfate is FeSO 4 . The sulfate ion is SO 4 2 - . The oxidation state of the sulfur is +6 (work it out!); therefore, the ion is more properly named the sulfate(VI) ion.

The sulfite ion is SO 3 2 - . The oxidation state of the sulfur is +4. This ion is more properly named the sulfate(IV) ion. The - ate ending indicates that the sulfur is in a negative ion.

FeSO 4 is properly named iron(II) sulfate(VI), and FeSO 3 is iron(II) sulfate(IV). Because of the potential for confusion in these names, the older names of sulfate and sulfite are more commonly used in introductory chemistry courses.

Using oxidation states to identify what has been oxidized and what has been reduced

This is the most common function of oxidation states. Remember:

In each of the following examples, we have to decide whether the reaction is a redox reaction, and if so, which species have been oxidized and which have been reduced.

Example \(\PageIndex{4}\):

This is the reaction between magnesium and hydrogen chloride:

\[ \ce{Mg + 2HCl -> MgCl2 +H2} \nonumber\]

Assign each element its oxidation state to determine if any change states over the course of the reaction:

The oxidation state of magnesium has increased from 0 to +2; the element has been oxidized. The oxidation state of hydrogen has decreased—hydrogen has been reduced. The chlorine is in the same oxidation state on both sides of the equation—it has not been oxidized or reduced.

Example \(\PageIndex{5}\):

The reaction between sodium hydroxide and hydrochloric acid is:

\[ NaOH + HCl \rightarrow NaCl + H_2O\]

The oxidation states are assigned:

None of the elements are oxidized or reduced. This is not a redox reaction.

Example \(\PageIndex{6}\):

The reaction between chlorine and cold dilute sodium hydroxide solution is given below:

\[ \ce{2NaOH + Cl_2 \rightarrow NaCl + NaClO + H_2O} \nonumber\]

It is probable that the elemental chlorine has changed oxidation state because it has formed two ionic compounds. Checking all the oxidation states verifies this:

Chlorine is the only element to have changed oxidation state. However, its transition is more complicated than previously-discussed examples: it is both oxidized and reduced. The NaCl chlorine atom is reduced to a -1 oxidation state; the NaClO chlorine atom is oxidized to a state of +1. This type of reaction, in which a single substance is both oxidized and reduced, is called a disproportionation reaction.

Using oxidation states to determine reaction stoichiometry

Oxidation states can be useful in working out the stoichiometry for titration reactions when there is insufficient information to work out the complete ionic equation. Each time an oxidation state changes by one unit, one electron has been transferred. If the oxidation state of one substance in a reaction decreases by 2, it has gained 2 electrons.

Another species in the reaction must have lost those electrons. Any oxidation state decrease in one substance must be accompanied by an equal oxidation state increase in another.

Example \(\PageIndex{1}\):

Ions containing cerium in the +4 oxidation state are oxidizing agents, capable of oxidizing molybdenum from the +2 to the +6 oxidation state (from Mo 2 + to MoO 4 2 - ). Cerium is reduced to the +3 oxidation state (Ce 3 + ) in the process. What are the reacting proportions?

The oxidation state of the molybdenum increases by 4. Therefore, the oxidation state of the cerium must decrease by 4 to compensate. However, the oxidation state of cerium only decreases from +4 to +3 for a decrease of 1. Therefore, there must be 4 cerium ions involved for each molybdenum ion; this fulfills the stoichiometric requirements of the reaction.

The reacting proportions are 4 cerium-containing ions to 1 molybdenum ion.

Here is a more common example involving iron(II) ions and manganate(VII) ions:

A solution of potassium manganate(VII), KMnO 4 , acidified with dilute sulfuric acid oxidizes iron(II) ions to iron(III) ions. In the process, the manganate(VII) ions are reduced to manganese(II) ions. Use oxidation states to work out the equation for the reaction.

The oxidation state of the manganese in the manganate(VII) ion is +7, as indicated by the name (but it should be fairly straightforward and useful practice to figure it out from the chemical formula)

In the process of transitioning to manganese(II) ions, the oxidation state of manganese decreases by 5. Every reactive iron(II) ion increases its oxidation state by 1. Therefore, there must be five iron(II) ions reacting for every one manganate(VII) ion.

The left-hand side of the equation is therefore written as: MnO 4 - + 5Fe 2 + + ?

The right-hand side is written as: Mn 2 + + 5Fe 3 + + ?

The remaining atoms and the charges must be balanced using some intuitive guessing. In this case, it is probable that the oxygen will end up in water, which must be balanced with hydrogen. It has been specified that this reaction takes place under acidic conditions, providing plenty of hydrogen ions.

The fully balanced equation is displayed below:

\[ MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+} \nonumber\]

Contributors and Attributions

Jim Clark ( Chemguide.co.uk )

What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results? K I – 3 H 2 S – – 4 O 6 F e – – – 3 O 4 C – – H 3 C H 2 O H C – – H 3 C – – O O H

K i – 3 in k i 3 , the oxidation number (o.n.) of k is + 1. hence, the average oxidation number of i is − 1 3 . however, o.n. cannot be fractional. therefore, we will have to consider the structure of k i 3 to find the oxidation states. in a k i 3 molecule, and atom of iodine forms a coordinate covalent bond with an iodine molecule. + 1 k + [ 0 i − 0 i ← − i i ] hence, in a ki3 molecule, the o.n. of the two i atoms forming the i 2 molecule is 0, whereas the o.n. of the i atom forming the coordinate bond is –1. h 2 s – – 4 o 6 + 1 h 2 x s o 4 − 2 o 6 now, 2 ( + 1 ) + 4 ( x ) + 6 ( − 2 ) = 0 ⇒ 2 + 4 x − 12 = 0 ⇒ 4 x = 10 ⇒ x = + 2 1 2 however, o.n. cannot be fractional. hence, s must be present in different oxidation states in the molecule. the o.n. of two of the four s atoms is +5 and the o.n. of the other two s atoms is 0. f e – – – 3 o 4 on taking the o.n. of o as –2, the o.n. of fe is found to be + 2 2 3 . however, o.n. cannot be fractional. here, one of the three fe atoms exhibits the o.n. of +2 and the other two fe atoms exhibit the o.n. of +3. + 2 f e o , + 3 f e 2 o 3 c – – h 3 c h 2 o h x c 2 + 1 h 6 − 2 o 2 ( x ) + 4 ( + 1 ) + 1 ( − 2 ) = 0 ⇒ 2 x + 6 − 2 = 0 ⇒ x = − 2 hence, the o.n. of c is - 2. x c 2 + 1 h 4 2 o 1 2 ( x ) + 4 ( + 1 ) + 2 ( − 2 ) = 0 ⇒ 2 x + 4 − 4 = 0 ⇒ x = 0 however, 0 is average o.n. of c. the two carbon atoms present in this molecule are present in different environments. hence, they cannot have the same oxidation number. thus, c exhibits the oxidation states of +2 and –2 in c h 3 c o o h ..

What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results?

(a) K I 3 (b) H 2 S 4 O 6 (c) Fe 3 O 4 (d) C H 3 C H 2 OH (e) C H 3 C OOH

Assign oxidation numbers to the underlined elements in each of the following species: N a H 2 P O 4

N a H S – – O 4

H 4 P – – 2 O 7

K 2 M n – –– – O 4

C a O – – 2

N a B – – H 4

H 2 S – – 2 O 7

K A l S – – O 4 ) 2 .12 H 2 O

Assign oxidation number to the underlined elements in each of the following species. C a O – – 2

Let x be the oxidation number of o in c a o 2 . 2 + 2 x = 0 x = − 1 hence, the oxidation number oxygen in cao 2 is -1.

Assign oxidation numbers to the underlined elements in each of the following species: N a H 2 P O 4

N a H S – – O 4

H 4 P – – 2 O 7

K 2 M n – –– – O 4

C a O – – 2

N a B – – H 4

H 2 S – – 2 O 7

K A l S – – O 4 ) 2 .12 H 2 O

Assign oxidation numbers to the underlined elements in each of the following species:

(a) NaH 2 P O 4 (b) NaH S O 4 (c) H 4 P 2 O 7 (d) K 2 Mn O 4

(e) Ca O 2 (f) Na B H 4 (g) H 2 S 2 O 7 (h) KAl( S O 4 ) 2 .12 H 2 O