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Questions  

Given ECr3+Cr=0.72V and  EFe2+Fe=0.42V The potential of the cell CrCr3+(0.1M)Fe2+(0.01M)Fe is

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By Expert Faculty of Sri Chaitanya
a
-0.26 V 
b
 0.26 V
c
0.339 
d
 -0.329
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detailed solution

Correct option is B

The cell potential is given by Ecell =EREL where 

Right half-cell  Fe2++2eFe

 ER=ERRT2Fln1Fe2+/c =0.42V+0.059V2log(0.01) =0.42V0.06V=0.48V 

Left half-cell  Cr3++3eCr

EL=ELRT3Fln1Cr3+/c=0.72+0.059V3log(0.1) =0.72V0.02V=0.74V

Hence, Ecell=0.48V(0.74V)=0.26V


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