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Question

The following reactions occurs in the cell Mg(s)+2Ag+(0.0001M)Mg2+(0.130M)+2Ag(s). Calculate E(cell)E0(cell)=3.17volt

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Solution

Ecell=E0cell0.0591nlogQ

Ecell=E0cell0.0591nlog[Mg+2][Ag+]2
Mg changes to Mg+2 by losing 2 electrons and 2Ag+ changes to 2Ag by gaining 2 electrons. So, exchange of 2 electrons takes place. Hence, n=2.
Ecell=3.170.05912log[0.130][0.0001]2
Ecell=3.17(0.02955×7.114) V
Ecell=3.170.21=2.96 V

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