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)

2,5 and 16

Answer 2

The half equations of the reaction are
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, MnO_4^- \rightarrow \, \, Mn^{2+}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, C_2O_4^{2-} \rightarrow \, CO_2$
The balanced half equations are
$\, \, \, MnO_4^- + 8H^{+} + 5e^- \rightarrow \, Mn^{2+} +H_2 O$
$\, \, \, \, \, \, \, \, \, \, \, \, \, C_2 O_4^{2-} \rightarrow \, \, 2 CO_2+2e^-$
On equating number of electrons, we get
$2MnO_4^- + 16H^+ +10e^- \rightarrow \, 2Mn^{2-}+8H_2O$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, 5C_2O_4^- \rightarrow \, \, 10CO_2 +10e^-$
On adding both the equations, we get
$2MnO_4^- + 5C_2O_4^-+16H^+ \rightarrow \, 2Mn^{2+}$
\hspace40mm +2 \times 5CO_2 +\frac{16}{2}H_2O
Thus .x, y and z are 2, 5 and 16 respectively