Question

Consider the following reaction,
$xMnO_4^- + yC_2O_4^{2-}+zH^+ \rightarrow xMn^{2+}+2yCO_2 + \frac{z}{2}H_2O$
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)

The half equations of the reaction are $ MnO _{4}^{-} \longrightarrow Mn ^{2+} $ $C _{2} O _{4}^{2-} \longrightarrow CO _{2}$ The balanced balf equations are $MnO _{4}^{-}+8 H ^{+}+5 e^{-} \longrightarrow Mn ^{2+}+4 H _{2} O$ $C _{2} O _{4}^{2-} \longrightarrow 2 CO _{2}+2 e^{-}$ On equating number of electrons, we get $2 MnO _{4}^{-}+16 H ^{+}+10 e^{-} \longrightarrow 2 Mn ^{2+}+8 H _{2} O$ $5 C _{2} O _{4}^{2-} \longrightarrow 10 CO _{2}+10 e^{-}$ On adding both the equations, we get $2 MnO _{4}^{-}+5 C _{2} O _{4}^{-}+16 H ^{+} \longrightarrow 2 Mn ^{2+} +2 \times 5 CO _{2}+\frac{16}{2} H _{2} O$ Thus $x, y$ and $z$ are $2,5$ and $16$ respectively.