2MnO₄ - + bC₂O₄²⁻ + cH + → xMn²⁺ + yCO₂ + zH₂O
If the above equation is balanced with integer coefficients, the value of c is ___________. (Round off to the Nearest Integer).

Solution

Chemistry ma’am · 4 steps

Balance the electrons first, then let the charges fix the number of H⁺ for you.

1. Oxidation states. Mn goes from +7 in MnO₄⁻ to +2 in Mn²⁺, so it gains 5 electrons. Each C goes from +3 in C₂O₄²⁻ to +4 in CO₂, and with two carbons each oxalate ion loses 2 electrons.
2. Match electrons. Two MnO₄⁻ gain 2 × 5 = 10 electrons. To lose 10 electrons you need b = 5 oxalate ions. Then x = 2 and y = 10.
3. Balance charge. Left side: 2(-1) + 5(-2) + c = c - 12. Right side: 2(+2) = +4. Setting them equal gives c - 12 = 4, so c = 16.

   2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

4. So. Hydrogen gives z = 8, and oxygen checks out (8 + 20 = 28 on both sides). The value of c is 16.

The common slip: 26. Counting 1 electron per oxalate ion instead of 2 (one per carbon) gives b = 10, and the charge balance then gives 26, not 16.

The idea to keep. In an acidic redox equation, balance the electrons first. After that, the H⁺ count is just whatever makes the total charge equal on both sides.

Answer from the NTA final answer key. Solution by PrepWiser.