Calculate the e.m.f. of the cell at `25^(@)C` `Cr|Cr^(3+)(0.1M)||Fe^(2+)(0.01M)|Fe`
Given `E_((Cr^(2+)//Cr))^(@)=-0.75V,E_((Fe^(2+)//Fe))^(@)=-0.44`

Explanation

At anode, `2Cr rarr 2Cr^(3+) + Fe`
At cathode,
`ul(3Fe^(2+) rarr 6e^(-) rarr 3Fe)`
`ul(2Cr + 3Fe^(2+) rarr 2Cr^(3+) + 3Fe)`
According to Nernest, `E= E^(@)- (0.059)/(6)"log" ([Cr^(3+)]^(2))/([Fe^(2+)]^(3))`
`E^(@)= -0.44+ 0.75`
=0.31V
`E= 0.31- (0.059)/(6) "log" ((0.1)^(2))/((0.01)^(3))`
`E= 0.31- (0.059)/(6) "log" ((0.01))/((0.000001))`
`=0.31 - (0.059)/(6) xx 4.007= 0.27V`