Question

Identify the correct balanced redox reaction by using ion-electron method:
$Bi{O}_3^{-}+M{n}^{2+}\to B{i}^{3+}+Mn{O}_4^{-}+H_{2}O$

• A. $5Bi{O}_3^{-}+22H^{+}+M{n}^{2+}\to 5B{i}^{3+}+7H_{2}O+Mn{O}_4^{-}$
• B. $5Bi{O}_3^{-}+14H^{+}+2M{n}^{2+}\to 5B{i}^{3+}+7H_{2}O+2Mn{O}_4^{-}$
• C. $2Bi{O}_3^{-}+4H^{+}+M{n}^{2+}\to 2B{i}^{3+}+2H_{2}O+Mn{O}_4^{-}$
• D. $6Bi{O}_3^{-}+12H^{+}+3M{n}^{2+}\to 6B{i}^{3+}+6H_{2}O+3Mn{O}_4^{-}$

Answer 1

The correct option is B $5Bi{O}_3^{-}+14H^{+}+2M{n}^{2+}\to 5B{i}^{3+}+7H_{2}O+2Mn{O}_4^{-}$

Reduction half cell :
$\stackrel{+5}{Bi}{O}_3^{-}\to B{i}^{3+}$
Oxidation half cell :
$M{n}^{2+}\to \stackrel{+7}{Mn}{O}_4^{-}$
Taking reduction half cell $Bi$ is balanced. So add $3H_{2}O$
in product side. Now to balance hydrogen 6 $H^{+}$ have to add in reactant side.
$Bi{O}_3^{-}+6H^{+}\to B{i}^{3+}+3H_{2}O$
Now balance the charge by adding $2e^{-}$ in reactant side:
$Bi{O}_3^{-}+6H^{+}+2e^{-}\to B{i}^{3+}+3H_{2}O$

For oxidation half cell :
$Mn$ is balanced. So add $4H_{2}O$
in reactant side. Now to balance hydrogen $8H^{+}$ have to add at product side.
$M{n}^{2+}+4H_{2}O\to Mn{O}_4^{-}+8H^{+}$
Now balance the charge by adding $5e^{-}$ at product side:
$M{n}^{2+}+4H_{2}O\to Mn{O}_4^{-}+8H^{+}+5e^{-}$

$Bi{O}_3^{-}+6H^{+}+2e^{-}\to B{i}^{3+}+3H_{2}O--\left(i\right)$

$M{n}^{2+}+4H_{2}O\to Mn{O}_4^{-}+8H^{+}+5e^{-}--\left(ii\right)$
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$\left(i\right)\times 5+\left(ii\right)\times 2,$ we get $14H^{+}+5Bi{O}_3^{-}+2M{n}^{2+}\to 5B{i}^{3+}+2Mn{O}_4^{-}+7H_{2}O$
Hence, (b) is the correct balanced reaction.