Question

Which of the following equations is a balanced one?

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


(i) $2e^{-}+6H^{+}+Bi{O}_3^{-}\to B{i}^{3+}+3H_{2}O\text{reduction}$
(ii) $4H_{2}O+M{n}^{2+}\to Mn{O}_4^{-}+8H^{+}+5e^{-}\text{oxidation}$
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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.