Question

Complete and balance the following reactions:

${Mn}^{2+}+Pb{O}_2⟶Mn{O}_4^{-}+H_{2}O$

• A. $2{Mn}^{2+}+5Pb{O}_2+4H^{+}\to 2Mn{O}_4^{-}+5{Pb}^{2+}+2H_{2}O$
• B. ${5Mn}^{2+}+2Pb{O}_2+6H^{+}\to 5Mn{O}_4^{-}+2{Pb}^{2+}+3H_{2}O$
• C. $2{Mn}^{2+}+5Pb{O}_2+4H^{+}\to 2Mn{O}_4^{-}+5{Pb}^{2+}+4H_{2}O$
• D. None of the above

Answer 1

The correct option is A $2{Mn}^{2+}+5Pb{O}_2+4H^{+}\to 2Mn{O}_4^{-}+5{Pb}^{2+}+2H_{2}O$

$P{b}^{4+}+M{n}^{2+}\to Mn{O}_4^{-}+P{b}^{2+}$

First consider two half reactions of oxidation and reduction,

$P{b}^{4+}+2e^{-}\to P{b}^{2+}$

$M{n}^{2+}\to M{n}^{7+}+5e^{-}$

Balance the electrons at both half cell reactions,

$5P{b}^{4+}+10e^{-}\to 5P{b}^{2+}$

$2M{n}^{2+}\to 2M{n}^{7+}+10e^{-}$

Add both the reactions and balance the oxygen using water and base,

$5Pb{O}_2+2M{n}^{2+}+4H^{+}\to 2Mn{O}_4^{-}+5P{b}^{2+}+2H_{2}O$

Hence,option A is correct.