How much charge is required for 1 mol of Al3 reduced to A12
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Answer:
Al3++3e−→Al
The reduction will require 3 Faraday of electricity or 3×96500=2.895×105C.
Cu2++2e−→Cu
The reduction will require 2 Faraday of electricity or 2×96500=1.93×105C.
MnO4−+5e−→Mn2+
The reduction will require 5 Faraday of electricity or 5×96500=4.825×105C.
Note: 1 mole of electron has a charge of 1 faraday.
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