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, `Cr rarr Cr^(3+) + 3e^(-) xx 2`
At cathode, `Fe^(2+) + 2e^(-) rarr Fe xx3`
`3Fe^(2+) + 6e^(-) rarr 3Fe`
`ul(2 Cr rarr 2Cr^(3+) + 2e^(-))`
`ul(2Cr+ 3Fe^(2+) rarr 2Cr^(3+) + 3Fe)`
`E= E^(@) - (0.059)/(6) "log. ([Cr^(3+)]^(2))/([Fe^(2+)]^(3))`
`E= 0.31- (0.059)/(6)"log" ((0.1)^(2))/((0.01)^(3))`
`=0.31- (0.059)/(6) xx 4.007= 0.27V`
`Delta G^(@)= -nE^(@)F= 2(0.357) (96500)`
`=68.9 kJ mol^(-1)`