Question

When the following oxidation–reduction reaction is balanced with lowest whole-number coefficients, what is the coefficient for Mn2+(aq)?

__MnO4–(aq) + __C2H5CHO(aq) + __H+ (aq) → __Mn2+(aq) + __C2H5COOH(aq) +__ H2O(l)

When I balanced this equation I got the following:

1MnO4–(aq) + 1C2H5CHO(aq) + 6H+ (aq) → 1Mn2+(aq) + 1C2H5COOH(aq) +3 H2O(l)

so therefore, the answer would be 1. However, the answer is 2. Can someone explain why the answer is 2 and what I did wrong?

Answer 1

Reduction half reaction:

MnO₄^-(aq) + 5 e^- -----> Mn²⁺(aq)

Mn = +7           Mn = +2

Balancing O by adding H2O

MnO₄^-(aq) + 5 e^- -----> Mn²⁺(aq) + 4 H₂O (l)

Balancing H by adding H⁺

MnO₄^-(aq) + 5 e^- + 8 H⁺ (aq) -----> Mn²⁺(aq) + 4 H₂O (l)                    ---------(1)

oxidation half reaction:

C₂H₅CHO(aq)       ------>     C₂H₅COOH + 2e^-

C =x+1-2 => +1                        C =x+1-(2*2) => +3

Balancing O by adding H2O

C₂H₅CHO(aq) + H₂O (l)      ------>     C₂H₅COOH + 2e^-

Balancing H by adding H+

C₂H₅CHO(aq) + H₂O (l)      ------>     C₂H₅COOH + 2e^- + 2H⁺ (aq) -----(2)

Add equation (1)*2 + (2)*5

2 MnO₄^-(aq) + 10 e^- + 16 H⁺ (aq) -----> 2 Mn²⁺(aq) + 8 H₂O (l)

5 C₂H₅CHO(aq) + 5 H₂O (l)      ------>   5 C₂H₅COOH + 10 e^- + 10 H⁺ (aq)

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2 MnO₄^-(aq) + 5 C₂H₅CHO(aq) + 6 H⁺(aq) ------> 2 Mn²⁺(aq) + 5 C₂H5COOH (aq) + 3 H₂O (l)

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