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Question

Calculate the emf of the cell.
Mg(s)|Mg2+(0.2 M)Ag+(1×103)|Ag
EAg+|Ag=+0.8 volt,EMg2+|Mg=2.37 volt
What will be the effect on emf if concentration of Mg2+ ion is decreased to 0.1 M

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Solution

ECell=ECathodeEAnode
=0.80(2.37)=3.17 volt
Cell reaction, Mg+2Ag+2Ag+Mg2+
Ecell=ECell0.0591nlogMg2+[Ag+]2
=3.170.05912log0.2[1×103]2
=3.170.1566=3.0134 volt
when Mg2+=0.1 M
Ecell=Ecell0.05912log0,1(1×103)2
=(3.170.1477)volt
=3.0223 volt.

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