Calculate the cell potential for the following cell:

Pb/Pb2+ (1x10-3 M) // Pt, 1/2 Cl2 (1 bar) / Cl- (0.1 M)

GIVEN: E0 Pb2+/Pb= -0.13V and E01/2 Cl2/Cl-= 1.358V

Here, cell reaction will be :
Pb+ 2Cl‚Äč- --> Pb+2 + Cl2 
E0cell = E0(RHS) - E0(LHS) = 1.358V-(-0.13V) =1.488 V
Since, here no. of electrons involved n =2
So, cell potential is calculated as follows:
Ecell=E0cell- (0.591/n) log [Cl-]/[Pb+2] = 1.488V-(0.591/2) log [0.1M]/[10-3M]
Solving we get, 
Ecell = 0.897V

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