135.
How much charge is required for the following reductions:
1 mol of MnO4– to Mn2+?
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
192 Views