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Given ECr+3/Cr0=-0.72V;EFe2+/Fe0=-0.42 VThe potential for the cell,Cr|Cr3+(0.1M||Fe2+(0.01M| Fe is

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a

– 0.339 V

b

– 0.26 V

c

0.26 V

d

0.339 V

answer is C.

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Detailed Solution

From the given SRP values Cr electrode acts as anode Fe electrode acts as cathode.Cell reaction is2Cr(s)+3Fe(aq)+2→2Cr(aq)+3+3Fe(s)Applying Nernst equatioinEcell=Eco-Eao-0.05916log[Cr+3]2Fe+23Ecell =(-0.42-(-0.72))-0.05916log[0.1]2[0.01]3Ecell =0.3-0.0596log104=0.26 V
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