Question

For the redox reaction,
$Mn{O}_4^{-}+C_2{O}_4^{2-}+H^{+}⟶M{n}^{2+}+C{O}_2+H_{2}O$
the correct coefficients of the reactants for the balanced equation are

	$Mn{O}_4^{-}$	$C_2{O}_4^{2-}$	$H^{+}$
a	$2$	$16$	$5$
b	$2$	$5$	$16$
c	$16$	$5$	$2$
d	$5$	$16$	$2$

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (B)

The given redox reaction is
$Mn{O}_4^{-}+C_2{O}_4^{2-}+H^{+}⟶M{n}^{2+}+C{O}_2+H_{2}O$
The reaction can be balanced by considering the following steps;
Step I Balance the atoms except $H$ and $O$.
$Mn{O}_4^{-}+C_2{O}_4^{2-}+H^{+}⟶M{n}^{2+}+2C{O}_2+H_{2}O$
Step II Write the oxidation number of each atom

Step III Cross multiply by change in oxidation number
$\stackrel{+7}{\mathrm{Mn}}{O}_4^{-}⟶M{n}^{2+};5e^{-}$ gain
$\stackrel{+3}{C_2}{O}_4^{2-}⟶\stackrel{+4}{2C{O}_2};2x1e^{-}$ loss
$2Mn{O}_4^{-}+5C_2{O}_4^{-}+H^{+}⟶2M{n}^{2+}+10C{O}_2+H_{2}O$
Step IV Balance oxygen by adding $H_{2}O$ on deficient site.
$2Mn{O}_4^{-}+5C_2{O}_4^{2-}+H^{+}⟶2M{n}^{2+}+10C{O}_2+8H_{2}O$
Step V Balance hydrogen
$2Mn{O}_4^{2-}+5C_2{O}_4^{2-}+16H^{+}⟶2M{n}^{2+}+10C{O}_2+8H_{2}O$
$\therefore$ The coefficients of the reactants, $Mn{O}_4^{-},C_2{O}_4^{2-}$ and $H^{+}$ are $2,5$ and $16$ , respectively.