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 reaction are:

• A. $Mn{O}_4^{-}=2,C_2{O}_4^{2-}=5,H^{+}=16$
• B. $Mn{O}_4^{-}=16,C_2{O}_4^{2-}=3,H^{+}=12$
• C. $Mn{O}_4^{-}=15,C_2{O}_4^{2-}=16,H^{+}=12$
• D. $Mn{O}_4^{-}=2,C_2{O}_4^{2-}=16,H^{+}=5$

Answer 1

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

Write the reaction in two half-reactions, oxidation half and reduction half as:

Reduction: $Mn{O}_4^{-}\to M{n}^{2+}$

Oxidation: ${C_2{O}_4}^{2-}\to C{O}_2$

Balance all atoms other than $O$ and $H$.

$Mn{O}_4^{-}\to M{n}^{2+}$

${C_2{O}_4}^{2-}\to 2C{O}_2$

Now balance the oxygen atoms by adding $H_{2}O$ molecules.

$Mn{O}_4^{-}\to M{n}^{2+}+4H_{2}O$

${C_2{O}_4}^{2-}\to 2C{O}_2$

Now balance hydrogen atoms by adding $H^{+}$ ions.

$Mn{O}_4^{-}+8H^{+}\to M{n}^{2+}+4H_{2}O$

${C_2{O}_4}^{2-}\to 2C{O}_2$

To balance the charge, add electrons to a more positive side to equal the less positive side of the half-reaction.

$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$

${C_2{O}_4}^{2-}\to 2C{O}_2+2e^{-}$

Now multiply oxidation half-reaction by 5 and reduction half-reaction by 2. Add both the reactions we will get

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

Hence, the correct option is $A$