Question

Balance the following reaction by ion electron method:
$KMn{O}_4+H_{2}S{O}_4+HCl\to K_{2}S{O}_4+MnS{O}_4+C{l}_2+H_{2}O$

• A. $2KMn{O}_4+10HCl+2H_{2}S{O}_4\to 5C{l}_2+2MnS{O}_4+8H_{2}O$
• B. $2KMn{O}_4+10HCl+3H_{2}S{O}_4\to 5C{l}_2+2MnS{O}_4+8H_{2}O$
• C. $2KMn{O}_4+10HCl+H_{2}S{O}_4\to 5C{l}_2+2MnS{O}_4+8H_{2}O$
• D. $2KMn{O}_4+HCl+3H_{2}S{O}_4\to 5C{l}_2+2MnS{O}_4+8H_{2}O$

Answer 1

The correct option is B $2KMn{O}_4+10HCl+3H_{2}S{O}_4\to 5C{l}_2+2MnS{O}_4+8H_{2}O$

Oxidaiton half: $2C{l}^{-}\to C{l}_2+2e^{-}$-----(1)
Reduction half:
Setp - I : $Mn{O}_4^{-}\to M{n}^{2+}$
Step - II : $Mn{O}_4^{-}\to M{n}^{2+}+4H_{2}O$
Step - III : $Mn{O}_4^{-}+8H^{+}\to M{n}^{2+}+4H_{2}O$
Step - IV : $Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$------(2)
Adding equation (1) $\times 5$ with equation (2) $\times$ 2
$[2C{l}^{-}\to C{l}_2+2e^{-}]\times 5$
$[Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O]\times 2$
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$2Mn{O}_4^{-}+10C{l}^{-}+16H^{+}\to 5C{l}_2+2M{n}^{2+}+8H_{2}O$

So the balanced equation is
$2KMn{O}_4+10HCl+3H_{2}S{O}_4\to 5C{l}_2+2MnS{O}_4+8H_{2}O$