Balancing Redox Equations

Just like any other chemical equation, a redox reaction must be balanced for the law of conservation of mass (matter) to be obeyed. Unlike other chemical equations, balancing redox equations goes beyond making sure that the atom counts of the respective elements on both sides of the equation are the same. It also involves the balancing of charges, because the fundamental process in redox reactions is the transfer of electrons. Now, talking about charges, before you think of balancing a redox equation, we must first learn how to change the given equation into an ionic form.

Let's consider the reaction between Zn and CuSO4 to form ZnSO4 and Cu as an example:

                        Zn(s) + CuSO4(aq) ---> ZnSO4(aq) + Cu(s) ……(i)

In the above reaction, zinc metal reacts with an aqueous solution of copper(II)tetraoxosulphate(VI) to form a mixture of aqueous solution of zinc tetraoxosulphate(VI) and copper metal. Since CuSO4 and ZnSO4 are in aqueous form, it means they can easily dissociate (ionize) in water to give cations (positive ions) and anions (negative ions) as shown:

                        CuSO4(aq) ---> (Cu2+)(aq) + (SO4--)(aq)

                        ZnSO4(aq) ---> (Zn2+)(aq) + (SO4--)(aq)

Now, let's substitute the above into equation (i)

                       Zn(s) + (Cu2+)(SO4--)(aq) ---> (Zn++)(SO4--)(aq) + Cu(s) ……(ii)

Cancel out any term that appears on both sides of the equation. In this case, it's (SO4--)

                       Zn(s) + (Cu2+)(aq) ---> (Zn2+)(aq) + Cu(s) ……(iii)

So, equation (iii) above is the ionic form of equation (i). It depicts that the reaction is also a  displacement reaction, where the Zn metal goes into solution and displaces (replaces) the Cu2+ as Zn2+.

Note that the zinc metal, Zn(s) and copper metal, Cu(s) are not ionized because they are in the solid state. Substances like insoluble solids, liquids and gases do not dissociate in water. So they are always represented as molecules in ionic equations. Hence, whenever you see substances with the state symbols as solid (s), gas (g) or liquid (l) in an equation, do not break them up when writing out the ionic equations.

Consider the reaction between iron(ii)chloride (FeCl2) and potassium tetraoxomanganate(VII),

                     FeCl2(aq) + KMnO4(aq) ---> FeCl3(aq) + KCl(aq) + MnCl2(aq)

Write the equation in the dissociation form:

         (Fe2+)(Cl-)(aq) + (K+)(MnO4-)(aq) ---> (Fe3+)(Cl-)(aq) + (K+)(Cl-)(aq) + (Mn2+)(Cl-)(aq)

Clear terms that appear on both sides of the equation, whose oxidation states do not change. In this case, they are K+ and Cl-. This implies that these two specie do not take part in the reaction, leaving behind

                        (Fe2+)(aq) + (MnO4-)(aq) ---> (Fe3+)(aq) + (Mn2+)(aq)

The above can then be balanced 'redoxically'. To achieve this, there are basic rules that must be followed, and in using these rules, the nature of the medium (neutral, acidic or alkaline) in which the reactions occur must be taken into consideration.

Basic Rules Guiding How To Balance Redox Equations

The best way to balance a redox equation is by considering the two half-equations involved in the reaction, i.e, the oxidation half-equation and the reduction half-equation.

...In Neutral or Acidic Medium
1. Identify the oxidizing agent and reducing agent by checking for any change in the oxidation numbers of elements other than oxygen and hydrogen
2. Deduce what their products should be (if not provided), otherwise move to the next step.
3. Write out the two half-equations.
4. Balance the atoms of elements other than O and H atoms using the appropriate coefficients.
5. Balance O by adding the appropriate number of H2O molecules to the side deficient of O.
6. Balance H by adding the appropriate number of H+ to the side deficient of H.
7. Balance the charges by adding the appropriate number of electrons  to:
(i) the right hand side (RHS) of the oxidation half-equation, (ii) the left hand side (LHS) of the reduction half-equation (bearing in mind that electrons are negatively charged)
8. Balance the electron loss in the oxidation half-equation with the electron gain in the reduction half-equation, because the total number of electrons lost MUST always equal the total number of electrons gained.
9. Add the two balanced half-equations and
(i) cancel out terms that appear on both sides of the equation, if they are equal; otherwise
(ii) find the difference between like terms and write the net value on the side of the equation with the bigger term.

Now, let's walk through the rules with this equation, which has already been expressed in the ionic form:

                        (Fe2+)(aq) +  (MnO4-)(aq) ---> (Fe3+)(aq) + (Mn2+)(aq)

Step 1:
Oxidizing agent - (MnO4-) because it is reduced. This is confirmed by the decrease in the oxidation state of Mn from +7 in (MnO4-) to +2 in (Mn2+)
Reducing agent - Fe2+ because it is oxidized as confirmed by the increase in the oxidation number of Fe from +2 in (Fe2+) to (Fe3+)

Step 2/3:
Since the products are provided, then:

                              Fe2+ ---> Fe3+ …………(oxidation half-equation)

                              MnO4- --->Mn2+ ………(reduction half-equation)

Step 4:
Taking the oxidation half-equation:

                             Fe2+ ---> Fe3+ (balanced in terms of Fe atoms count)

Considering the reduction half-equation:

                           (MnO4-) ---> Mn2+ (balanced in terms of Mn atoms count)

Step 5/6:
Balancing O atoms
We will perform these steps to the reduction half-equation only because it is the only half-equation that contains oxygen

                            (MnO4-) ---> (Mn2+) + 4H2O

By balancing O with H2O, we have introduced H into the equation and that will be balanced thus:

                          (MnO4-) + 8(H+) ---> (Mn2+) + 4H2O

Step 7:
Balancing the charges
For the oxidation half-equation:

                                     Fe2+ ---> Fe3+

Net charge on LHS = +2 x 1 = +2
Net charge on RHS = +3 x 1 = +3
We need -1 to make RHS = LHS, and that is equivalent to 1e- . So, the balanced oxidation half-equation is thus:

                                    Fe2+ ---> Fe3+  +  e- …………(i)

For the reduction half-equation:

                                 (MnO4-) + 8(H+) ---> (Mn2+) + 4H2O

Net charge on LHS = (-1 x 1) + (+1 x 8) = -1 + 8 = +7
Net charge on RHS = (+2 x 1) + (0 x 4) = +2 + 0 = +2
Now, we need -5 to make LHS = RHS, and that is equivalent to 5e-. Hence, the balanced reduction half-equation will be:

                                (MnO4-) + 8(H+) + 5e-  ---> (Mn2+) + 4H2O …………(ii)

Step 8:
From equations (i) and (ii), we can see that 1e- is gained in the reduction half-equation against 5e- that are lost in the oxidation half-equation. Clearly, there is a deficit of 4e- that have to be accounted for. Hence, the need to balance the transfer of electrons.

To achieve this, multiply equation (i) by 5 and equation (ii) by 1

                                5(Fe2+) ---> 5(Fe3+)  +  5e- ………(iii)
                                (MnO4-) + 8(H+) + 5e-  ---> (Mn2+) + 4H2O …………(ii)

Step 9:
                   5(Fe2+) + (MnO4-) + 8(H+) + 5e-  --> 5(Fe3+) +  (Mn2+) + 4H2O + 5e- ………(iv)

Cancel out like terms (5e-), and the balanced overall redox equation will be:

                              5(Fe2+) + (MnO4-) + 8(H+) ---> 5(Fe3+) +  (Mn2+) + 4H2O ………(v)

Hint:
The fastest way of confirming whether the redox equation is balanced or not is to ensure the net charges on both sides of the equation are the same.
So, let's check:
Net charge on LHS = (+2 x 5) + (-1 x 1) + (+1 x 8) = +10 - 1 + 8 = +17
Net charge on RHS = (+3 x 5) + (+2 x 1) + (0 x 4) = +15 + 2 + 0 = +17
Correct!

Note:
1) From the set of equations in Step 9 above, it means that 5 moles of electrons are involved or transferred in the redox reaction between Fe2+ and MnO4-
2) The H+ (hydrogen ion) in the overall balanced equation indicates that the reaction occurred in acidic medium.
3) Whether in neutral or acidic medium, the result is the same. As seen in our example, the question was provided without the H+, which means it is a neutral medium. However, in the course of balancing the equation, we end up with the H+

...In Alkaline Medium
In addition to Rules 1 - 9 above, the following rules are necessary when balancing redox equations in alkaline medium:
10. Balance the H2O molecules in the overall equation obtained from Step 9 above, by adding 2 moles of OH- (hydroxyl ion) to both sides of the equation for every 1 molecule of H2O.
11. Combine the H+ and OH- that occur on the same side of the equation to form H2O molecules.
12. Go back to Rule 9(ii)

Now, using our example above, let's try to balance it in alkaline medium.

Steps 1 - 9 (remain the same)

Step 10:
Balancing H2O molecules

                             5(Fe2+) + (MnO4-) + 8(H+) ---> 5(Fe3+) +  (Mn2+) + 4H2O

There are 4 molecules of H2O on the RHS, and if we need 2OH- for every molecule of H2O, it implies that we will need to add 4*2OH- = 8OH- on both sides of the equation.

         5(Fe2+) + (MnO4-) + 8(H+) + 8(OH-)---> 5(Fe3+) +  (Mn2+) + 4H2O + 8(OH-) ………(vi)

Step 11:
Combining H+ and OH- to form H2O:
From equation (vi) above, if we combine the 8(H+) and 8(OH-) that occur on the LHS, it will be:
                                      
                            8H+  +  8OH-  ---> 8H2O

(Remember H+ + OH- ----> H2O)

Therefore, equation (vi) becomes:

                      5(Fe2+) + (MnO4-) + 8H2O ---> 5(Fe3+) +  (Mn2+) + 4H2O + 8(OH-) ………(vii)

Step 12:
Net off like terms, (i.e, H2O) and get the overall equation

                     5(Fe2+) + (MnO4-) + 4H2O ---> 5(Fe3+) +  (Mn2+) + 8(OH-) ………(viii)

Check:
Net charge on LHS = (+2 x 5) + (-1 x 1) + (0 x 4) = +10 - 1 + 0 = +9
Net charge on RHS = (+3 x 5) + (+2 x 1) + (-1 x 8) = +15 + 2 - 8 = +9
Correct!

Note:
The presence of the OH- in the last equation indicates that it is an alkaline medium.
Now, compare equation (v) to equation (viii). What is the difference?

Let's try and balance one more equation in both acidic and alkaline medium.

Question:
Write a balanced ionic equation for the redox reaction between
(a) acidified potassium heptaoxodichromate(vi) (K2Cr2O7) and hydrogen sulphide gas (H2S)
(b) in alkaline medium

Answer:
a) Since the reaction is between acidified K2Cr2O7 and H2S, it means it occurs in an acidic medium. So, let's fly:

                       K2Cr2O7(aq) + H2S(aq) ---> K2S(aq) + Cr2S3(s) + S(s) + H2O

                      2(K+)(Cr2O7--) + 2(H+)(S2-) ---> 2(K+)(S2-) + Cr2S3 + S + H2O

                      (Cr2O7--) + 2(H+) + (S2-) ---> Cr2S3 + S + H2O

Step 1:
In the above redox reaction, (Cr2O7--) is the oxidizing agent, because it is the specie that is reduced. The oxidation state of Cr has decreased from +6 in (Cr2O7--) to +3 in (Cr2S3). Likewise, the oxidation state of S2- has increased from -2 in (S2-) to 0 in S, which makes it the reducing agent. The H+ in the equation indicates that it is an acidic medium.
Note that chromium(iiii)sulphide, Cr2S3, is insoluble in water that is why it does not dissociate.

Steps 2/3:
                          (S2-) ---> S (oxidation half-equation)
                          (Cr2O7--)  ---> Cr2S3 (reduction half-equation)

Step 4:
                           (S2-) ---> S
                           (Cr2O7--) + 3S ---> Cr2S3

Steps 5/6:
                          (Cr2O7--) + 3S ---> Cr2S3 + 7H2O
                           (Cr2O7--) + 3S + 14H+ ---> Cr2S3 + 7H2O

Step 7:
                           (S2-) ---> S + 2e- …………(i)
                           (Cr2O7--) + 3S + 14H+ 12e- ---> Cr2S3 + 7H2O …………(ii)

Step 8:
To balance electrons, multiply eqn (i) by 6, and eqn (ii) by 1

                          6(S2-) ---> 6S + 12e-
                          (Cr2O7--) + 3S + 14H+ 12e- ---> Cr2S3 + 7H2O

Step 9:
                         6(S2-) + (Cr2O7--) + 3S + 14H+ 12e- ---> 6S + Cr2S3 + 7H2O + 12e-

Clear like terms

                        6(S2-) + (Cr2O7--) + 14H+  ---> 3S + Cr2S3 + 7H2O ………(iii)

Check:
Net charge on LHS = (-2 x 6) + (-2 x 1) + (1 x 14) = -12 - 2 + 14 = 0
Net charge on RHS = (0 x 3) + (0 x 1) + (0 x 7) = 0 + 0 + 0 = 0
Correct!

b) In alkaline medium

Step 10:
                      6(S2-) + (Cr2O7--) + 14H+ 14OH- ---> 3S + Cr2S3 + 7H2O + 14OH-

Step 11:
                      6(S2-) + (Cr2O7--) + 14H2O  ---> 3S + Cr2S3 + 7H2O + 14OH-

Step 12:
                      6(S2-) + (Cr2O7--) + 7H2O  ---> 3S + Cr2S3 + 14OH-

Check:
Net charge on LHS = (-2 x 6) + (-2 x 1) + (0 x 7) = -12 - 2 + 0 = -14
Net charge on RHS = (0 x 3) + (0 x 1) + (-1 x 14) = 0 + 0 - 14 = -14
Correct!

Do These:
1) What is the value of n in the following equation?
(Cr2O7--) + 14H+ + ne- ---> 2Cr3+ + 7H2O
A. 2    B. 3    C. 6    D. 7

2) Balance the following redox reaction:
MnO4- +  I- +  H+ ---> Mn2+  +  I2

3) What is the function of manganese (IV) oxide in the reaction represented by the following equation:
MnO2 + Cl- ---> (Mn2+) + Cl2 + H2O
Hence or otherwise, balance the equation.

4) Consider the redox reaction as represented by the following equation:
I2 + S2O3--  ---->  I- + S4O6--
Determine the number of moles of electrons transferred in this reaction

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