Question

In the following balanced reaction,
$XMn{O}_4^{-}+Y{C}_2{O}_4^{2-}+Z{H}^{+}\rightleftharpoons XM{n}^{2+}+2TC{O}_2+\frac{Z}{2}{H}_{2}O$
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)

$XMn{O}_4^{-}+Y{C}_2{O}_4^{2-}+Z{H}^{+}\rightleftharpoons XM{n}^{++}+2YC{O}_2+\frac{Z}{2}{H}_{2}O$
First half reaction
$Mn{O}_4^{-}\to M{n}^{++}....\left(i\right)$
On balancing
$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{++}+4H_{2}O....\left(ii\right)$
Second half reaction
$C_2{O}_4^{--}\to 2C{O}_2....\left(iii\right)$
On balancing
$C_2{O}_4^{--}\to 2C{O}_2+2e^{-}....\left(iv\right)$
On multiplying eqn. $\left(ii\right)$ by 5 and $\left(iv\right)$ by 2 and then adding we get
$2Mn{O}_4^{-}+5C_2{O}_4^{-}+16H^{+}\to 2M{n}^{++}+10C{O}_2+8H_{2}O$