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Electrochemistry

Question
CBSEENCH12005974
Wired Faculty App

How much charge is required for the following reductions:
1 mol of MnO4– to Mn2+?

Solution
MnO4-(aq) + 8H++5e-     →        Mn2+(aq) + 6H2O. 1 mol                            5 moles             1 mol

Formula required charge n × F
n = difference of charge on ions   
F is constant and equal to 96487 Coulombs

Charge on Mn in MnO4–
Charge on Oxygen is – 2
Mn + 4O     = – 1
Mn +4(–2)   = – 1
Mn              = +7
So our reaction is
MN 7+ à Mn 2+
n = 7– 2  = 5  
Here n= 5
Required charge will  = 5 × 96487 Coulombs
                                = 482435 Coulombs
                                = 4.82 × 105 Coulombs

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