Consider the following reaction: $$\mathrm{xMnO}_{4}^{-}+\mathrm{yC}_{2} \mathrm{O}_{4}^{2-}+\mathrm{zH}^{+} \longrightarrow \mathrm{xMn}^{2+}+2 \mathrm{yCO}_{2}+\frac{\mathrm{z}}{2} \mathrm{H}_{2} \mathrm{O}$$ The values of $\mathrm{x}, \mathrm{y}$ and z in the reaction are, respectively:

JEE Main · 2013 · 07 Apr 2013 · Offline · Redox Reactions and Electrochemistry

Consider the following reaction:

xMnO_{4}^{-} + yC_{2} O_{4}^{2 -} + zH^{+} ⟶ xMn^{2 +} + 2 yCO_{2} + \frac{z}{2} H_{2} O

The values of

x, y

and z in the reaction are, respectively:

Answer: (B)

Balance the two half-reactions. Oxalate is oxidised,

C X_{2} O X_{4} X^{2 -} 2 CO X_{2} + 2 e X^{-}

, and permanganate is reduced,

MnO X_{4} X^{-} + 8 H X^{+} + 5 e X^{-} Mn X^{2 +} + 4 H X_{2} O

. To set five electrons lost against five gained, multiply the oxalate half by

5

and the permanganate half by

2

, giving

2 MnO X_{4} X^{-} + 5 C X_{2} O X_{4} X^{2 -} + 16 H X^{+} - > 2 Mn X^{2 +} + 10 CO X_{2} + 8 H X_{2} O

. Reading off the coefficients,

x = 2

y = 5

and

z = 16

JEE Main · 2013 · 07 Apr 2013 · Offline · Single correct option