Question

By using the half-reaction method of electron exchange balance the redox reaction :
$KMn{O}_4+H_{2}S{O}_3\to K_{2}S{O}_4+MnS{O}_4+H_{2}S{O}_4+H_{2}O$
Which of the following partial equations is the correct reduction half-reaction for the balanced equation?

• A. $5S{O}_3^{2-}+5H_{2}O\to 5S{O}_4^{2-}+10H^{+}+10e^{-}$
• B. $2Mn{O}_4^{-}+16H^{+}+10e^{-}\to 2M{n}^{2+}+8H_{2}O$
• C. $S{O}_3^{2-}\to S{O}_4^{2-}+2e^{-}$
• D. $S{O}_3^{2-}+2H^{+}\to S{O}_4^{2-}+H_{2}O+2e^{-}$
• E. $M{n}^{7+}\to M{n}^{2+}+5e^{-}$

Answer 1

Answer: (B)

$2Mn{O}_4^{-}+16H^{+}+10e^{-}\to 2M{n}^{2+}+8H_{2}O$

The oxidation and reduction half of the above reaction are as follows:

Oxidation half

${S{O}_3}^{2-}\to {S{O}_4}^{2-}+2e^{-}$

Reduction half

${2Mn{O}_4}^{-}+16H^{+}+10e^{-}\to 2M{n}^{2+}+2e^{-}$

Hence B is the correct answer