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Question

An excess of liquid mercury is added to an acidified solution of 1.0×103 M Fe3+. It is found that 5 % of Fe3+ remains that the only reaction that occurs is:

2Hg+2Fe2+Hg2+2+2Fe2+. (write your answer to nearest integer)
Given : EFe3+/Fe2+ = 0.77 V.

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Solution

2Hg+2Fe3+Hg2+2+2Fe2+

Fe3+ ion concentration originally = 103M

=5100×103=5×105M

Fe3+ converted into Fe2+ ion

= (1.0×103)(5×105)=0.95×103

Hg2+2 ions at eq = 0.95×1032

Ecell=Ecell0.059nlog[Hg2+2][Fe2+]2][Fe3+]2

At equivalent, Ecell=0

0=E0.05912log0.95×1032[0.95×103]2[5×105]2

Ecell=0.05912log[95]310550=0.0276

Ecell=EFe3+IFe2+EHg2+2IHg

0.0276=0.77EHg2+2IHg

EHg2+IHg=0.77+0.0276

= 0.7976 V

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