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 respectively $Mn{O}_{4,}^{-}+C_2{O}_4^{-}+H^{+}:$

• A. $15,16,12$
• B. $16,3,12$
• C. $2,16,5$
• D. $2,5,16$

Answer 1

The correct option is D $2,5,16$

$MnO{{}_4}^{-}\left(aq\right)+C_{2}O{{}_4}^{2-}\left(aq\right)\to M{n}^{2+}\left(aq\right)+C{O}_2\left(g\right)$

Step 1 : Divide into two half - reactions.
$MnO{{}_4}^{-}\to M{n}^{2+}$
$C_{2}O{{}_4}^{2-}\to C{O}_2$

Step 2 : Balance main element
$MnO{{}_4}^{-}\to M{n}^{2+}$
$C_{2}O{{}_4}^{2-}\to 2C{O}_2$

Step 3 : Balance the $O$ atoms by adding $H_{2}O$
$MnO{{}_4}^{-}\to M{n}^{2+}+4H_{2}O$
$C_{2}O{{}_4}^{2-}\to 2C{O}_2$

Step 4 : Balance the $H$ atoms by a adding $H^{+}$
$8H^{+}+MnO{{}_4}^{-}\to M{n}^{2+}+4H_{2}O$
$C_{2}O{{}_4}^{2-}\to 2C{O}_2$

Step 5 : Balance charge with electrons

$5e^{-}+8H^{+}+\stackrel{+7}{Mn}O{{}_4}^{-}\to M{n}^{2+}+4H_{2}O...\left(i\right)$

$C_{2}O{{}_4}^{2-}\to 2C{O}_2+2e^{-}...\left(ii\right)$
On multiplying equation (i) by $2$ and equation (ii) by $5$, we get,
$2MnO{{}_4}^{-}+5C_{2}O{{}_4}^{2-}+16H^{+}\to 2M{n}^{2+}+10C{O}_2+8H_{2}O$.