Question

In the following balanced reaction,
$X \,MnO^{-}_{4} +YC_{2}O_{4}^{2-} +Z H^{+} {\rightleftharpoons} X\,Mn^{2+}+2TCO_{2} + \frac{Z}{2} H_{2}O$
values of X, Y and Z respectively are

• A. 2, 5, 16
• B. 8, 2, 5
• C. 5, 2, 16
• D. 5, 8, 4

Answer 1

Answer: (A)

$X \,MnO^{-}_{4} +YC_{2}O_{4}^{2-} +Z H^{+} {\rightleftharpoons} X\,Mn^{++}+2YCO_{2} + \frac{Z}{2} H_{2}O$
First half reaction
$MnO^{-}_{4}\to Mn^{++}\quad\quad.... \left(i\right)$
On balancing
$MnO^{-}_{4} + 8H^{+} + 5e^{-} \to Mn^{++}+4H_{2}O\quad\quad.... \left(ii\right)$
Second half reaction
$C_{2}O_{4}^{--} \to 2CO_{2}\quad\quad....\left(iii\right)$
On balancing
$C_{2}O^{--}_{4} \to 2CO_{2} + 2e^{-}\quad\quad....\left(iv\right)$
On multiplying eqn. $\left(ii\right)$ by 5 and $\left(iv\right)$ by 2 and then adding we get
$2MnO^{-}_{4} + 5C_{2}O^{-}_{4} + 16H^{+} \to 2Mn^{++} + 10CO_{2} + 8H_{2}O$