Question

Consider the following reaction,

$xMn{O}_4^{-}+y{C}_2{O}_4^{2-}+z{H}^{+}\to xM{n}^{2+}+2yC{O}_2+\frac{z}{2}{H}_{2}O$

The values of $x$, $y$ and $z$ in the reaction are, respectively:

• A. $5,2$ and $16$
• B. $2,5$ and $8$
• C. $2,5$ and $16$
• D. $5,2$ and $8$

Answer 1

Answer: (C)

$2,5$ and $16$

Let

$M{n}^{7+}\to M{n}^{2+}$ [Reduction]

and

$C^{3+}\to C^{4+}$ [Oxidation]

balance the equation

$M{n}^{7+}+5e^{-}\to M{n}^{2+}$ ---- (a)

$C_2^{3+}\to 2C^{4+}+2e^{-}$ ---- (b)

Multiply $2$ by equation $a$ and multiply $5$ by $b$ and adding them gives.

$2M{n}^{7+}+5C_2^{3+}\to 2M{n}^{2+}+10C^{4+}$

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

Balancing O and H considering acidic medium

$2Mn{O_4}^{-}+5C_2{O_4}^{2-}+16H^{+}\to 2M{n}^{2+}+10C{O}_2+8H_{2}O$

Hence

$x=2,$ $y=5,$ $z=16$