Question

For the redox reaction, $Mn{O}_4^{-}+C_2{O}_4^{2-}+H^{+}\to M{n}^{2+}+C{O}_2+H_{2}O$, the correct coefficients of the reactants for the balanced equation are _____________.

• A. $Mn{O}_4^{-}=2,C_2{O}_4^{2-}=16,H^{+}=5$
• B. $Mn{O}_4^{-}=16,C_2{O}_4^{2-}=5,H^{+}=2$
• C. $Mn{O}_4^{-}=5,C_2{O}_4^{2-}=16,H^{+}=2$
• D. $Mn{O}_4^{-}=2,C_2{O}_4^{2-}=5,H^{+}=16$

Answer 1

The correct option is D $Mn{O}_4^{-}=2,C_2{O}_4^{2-}=5,H^{+}=16$

The unbalanced redox reaction is $Mn{O}_4^{-}+C_2{O}_4^{2-}+H^{+}\to M{n}^{2+}+C{O}_2+H_{2}O$
Balance C atoms.

$Mn{O}_4^{-}+C_2{O}_4^{2-}+H^{+}\to M{n}^{2+}+2C{O}_2+H_{2}O$

The oxidation number of Mn decreases from $+7$ to $+2$.
The decrease in the oxidation number is $7-2=5$
The oxidation number of C increases from $+3$ to $+4$.
The increase in the oxidation number for 1 C atom is $4-3=1$
The increase in the oxidation number for 2 C atom is $2\times 1=2$

To balance the increase in the oxidation number with decrease in the oxidation number, multiply Mn containing species with 2 and C containing species with 5.

$2Mn{O}_4^{-}+5C_2{O}_4^{2-}\to 2M{n}^{2+}+10C{O}_2$

Balance O atoms by adding 8 water molecules to the products side.

$2Mn{O}_4^{-}+5C_2{O}_4^{2-}\to 2M{n}^{2+}+10C{O}_2+8H_{2}O$

Balance H atoms by adding 16 $H^{+}$ ions to the reactants side.

$2Mn{O}_4^{-}+5C_2{O}_4^{2-}+16H^{+}\to 2M{n}^{2+}+10C{O}_2+8H_{2}O$

This is the balanced equation.