Transition and InnerTransition Elements - Result Question 96
####23. Consider the following reaction,
(2013 Main) $x MnO _4^{-}+y C _2 O _4^{2-}+z H^{+} \longrightarrow x Mn^{2+}+2 y CO _2+\frac{z}{2} H _2 O$ 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
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Solution:
- The half equations of the reaction are
$$ \begin{aligned} & MnO _4^{-} \longrightarrow Mn^{2+} \ & C _2 O _4^{2-} \longrightarrow CO _2 \end{aligned} $$
The balanced half equations are
$$ \begin{aligned} MnO _4^{-}+8 H^{+}+5 e^{-} & \longrightarrow Mn^{2+}+4 H _2 O \ C _2 O _4^{2-} & \longrightarrow 2 CO _2+2 e^{-} \end{aligned} $$
On equating number of electrons, we get
$$ \begin{aligned} 2 MnO _4^{-}+16 H^{+}+10 e^{-} & \longrightarrow 2 Mn^{2+}+8 H _2 O \ 5 C _2 O _4^{2-} & \longrightarrow 10 CO _2+10 e^{-} \end{aligned} $$
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.
