Question

In the balanced equation, the sum of coefficient of $H^{+}$ and $C{O}_2$ is:
${Mn{O}_4}^{-}+C_2{O}_4^{2-}+H^{+}\to M{n}^{2+}+C{O}_2+H_{2}O$

• A. $16$
• B. $8$
• C. $4$
• D. $26$

Answer 1

Answer: (D)

$26$

To balance the atoms other than $H$ and $O$, coefficient $2$ is added in front of $C{O}_2$.

$C$ is oxidized. Its oxidation number increases from $+3$ to $+4$. The net increase in the oxidation number is $+2$.

$Mn$ is reduced. The oxidation number decreases from $+7$ to $+2$. The net decrease in the oxidation number is $+5$.

Hence, the coefficient of $C$ is multiplied by $5$ and the coefficient of $Mn$ is multiplied by $2$.

To balance oxygen atoms, $7$ water molecules are added to the right. To balance $H$ atoms, $16$ protons are added to left.

Hence, the balanced reaction is $2Mn{O}_4^{-}+5C_2{O}_4^{2-}+16H^{+}\to 10C{O}_2+8H_{2}O+2M{n}^{2+}$

The sum of coefficients of $H^{+}$ and $C{O}_2$ is $26$.