Construct galvanic cells from the following pairs of half cells and calculate their emf at `25^@C`
`Fe^(3+)(0.1M), Fe^(2+) (1M) (Pt) E_(Fe^(3+),Fe^(2+))=0.77`volt
`AgCl(s),Cl^- (0.001M) |Ag E_(AgCl,Cl^-)^@=0.22` volt

Explanation

For the half cell reaction
`Fe^(3+)+e leftrightarrow Fe^(2+)`,
`E_(Fe^(3+),Fe^(2+))=E_(Fe^(3+),Fe^(2+))^@-0.0591/1 log""([Fe^(2+)])/([Fe^(3+)])`
`=0.77- 0.0591/1 log"" 1/0.1=0.71` volt
And, for the half cell reaction
`AgCl(s)+ e=Ag(s)+Cl^(-)`