Question

how to balance this redox reaction? MnO4 + C2O4 2 + H+ → Mn+2 + CO2 + H2O

Answer 1

2MNO4+5C2O4+16H=2MN+10Co2+8H2O

Answer 2

Answer:

To balance a redox reaction:

1. Write down the skeleton unbalanced equation:

$\mathrm{MnO}_{4}+\mathrm{C}_{2} \mathrm{O}_{4}(2)+\mathrm{H}+\rightarrow \mathrm{Mn}+2+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}$

2. Separating the reaction into half as it is a combination of reduction and oxidation:

Oxidation reaction: $C^{+3} 2 O^{-2} 4^{2} \rightarrow 2 C^{4} O^{-2} 2+2 e^{-}$

Reduction reaction: $\mathrm{Mn}^{+7} \mathrm{O}^{-2} 4+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{+2+}$

3. Balancing atoms in each half of the reaction:

Oxidation: $\mathrm{C}^{+3} 2 \mathrm{O}^{-2} \mathrm{4}^{2-} \rightarrow 2 \mathrm{C}^{+4} \mathrm{O}^{-2} 2+2 \mathrm{e}^{-}$

Reduction: $\mathrm{Mn}^{+7} \mathrm{O}^{-2} 4+5 \mathrm{e}^{-}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{+2+}+4 \mathrm{H}_{2} \mathrm{O}$

4. Equalizing the electron lost to electron gain:

Oxidation: $5 \mathrm{C}^{+3} 2 \mathrm{O}^{-2} 4^{2-} \rightarrow 10 \mathrm{C}^{+4} \mathrm{O}^{-2} 2+2 \mathrm{e}$  

Reduction: $2 \mathrm{Mn}^{+7} \mathrm{O}^{-2} 4+10 \mathrm{e}^{-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{+22+}+8 \mathrm{H}_{2} \mathrm{O}$

5. Adding both the half reactions:

$5 \mathrm{C}^{+3} 2 \mathrm{O}^{-2} \mathrm{H}^{2-}+2 \mathrm{Mn}^{+7} \mathrm{O}^{-2} 4+10 \mathrm{e}^{-}+16 \mathrm{H}^{+} \rightarrow 10 \mathrm{C}^{+4} \mathrm{O}^{-2} 2+2 \mathrm{Mn}^{+2+}+8 \mathrm{H} 2 \mathrm{O}+1 \mathrm{Oe}$

6. Upon simplifying the equation:

$5 \mathrm{C}^{+3} 2 \mathrm{O}^{-2} \mathrm{4}^{2-}+2 \mathrm{Mn}^{+7} \mathrm{O}^{-2} 4+16 \mathrm{H}^{+} \rightarrow 10 \mathrm{C}^{4} \mathrm{O}^{-2} 2+2 \mathrm{Mn}^{+2+}+8 \mathrm{H}_{2} \mathrm{O}$

7. Balancing the charges on both sides, it is 4 on each side:

8. Final balanced equation:

$5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-}+2 \mathrm{MnO}_{4}^{-}+16 \mathrm{H}^{+} \rightleftharpoons 10 \mathrm{CO}_{2}+2 \mathrm{Mn}^{2+}+8 \mathrm{H}_{2} \mathrm{O}$