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NCERT Class XII Chemistry
Chapter - Electrochemistry
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Question : 12 of 32
Marks: +1, -0
How much charge is required for the following reductions:
(i) 1 mol of Al3+\mathrm{Al}^{3+} to Al\mathrm{Al}?
(ii) 1 mol of Cu2+\mathrm{Cu}^{2+} to Cu\mathrm{Cu} ?
(iii) 1 mol of MnO4−\mathrm{MnO}_4^{-} to Mn2+\mathrm{Mn}^{2+} ?
Solution:  
(i) The given reaction is
Al3+1 mole+3e−3 mole→Al\underset{\text{1 mole}}{\mathrm{Al}^{3+}} + \underset{\text{3 mole}}{3\mathrm{e}^{-}} \rightarrow \mathrm{Al}
∴ 3 mole electrons are needed for reduction of 1 mole of Al3+\mathrm{Al}^{3+} to Al\mathrm{Al}
3 mole electrons = 3 Faraday
= 3 × 96500 coulombs
= 2.895 × 105^{5}coulombs
(ii) The given reaction is
Cu2+1 mole+2e−2 mole→Cu\underset{\text{1 mole}}{\mathrm{Cu}^{2+}} + \underset{\text{2 mole}}{2\mathrm{e}^{-}} \rightarrow \mathrm{Cu}
∴ 2 mole electrons are needed for reduction of 1 mole of Cu2+\mathrm{Cu}^{2+} to Cu\mathrm{Cu}
2 mole electrons = 2 Faraday
= 2 × 96500 coulombs
= 1.93 × 105^{5}coulombs
(iii)The given reaction is
MnO4−(aq)1 mole+8H++5e−5 mole→Mn2+(aq)+4H2O(l)\underset{\text{1 mole}}{\mathrm{MnO}_4^{-}(\mathrm{aq})} + 8\mathrm{H}^{+} + \underset{\text{5 mole}}{5\mathrm{e}^{-}} \rightarrow \mathrm{Mn}^{2+}(\mathrm{aq}) + 4\mathrm{H}_2\mathrm{O}(\mathrm{l})
∴5 mole electrons are needed for reduction of 1 mole of MnO4−\mathrm{MnO}_4^{-} to Mn2+.\mathrm{Mn}^{2+}.
5 mole electrons = 5 Faradays
= 5 × 96500 coulombs
= 4.825 × 105^{5}coulombs
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