Correct Answer - D
`I^(-)` get oxidized to `I_(2)` hence will form anode and `Cr_(2)O_(7)^(2-)` get reduced to `Cr^(3+)` hence will form cathode.
`E_(cell)^(o)=E_("cathode")^(o)-E_("anode")^(o),E_(cell)^(o)=E_(Cr_(2)O_(7)^(-2))-E_(I_(2))^(o)`.
`0.79=1.33-E_(I_(2))^(o),E_(I_(2))^(o)=1.33-0.79,E_(I_(2))^(o)=0.54V`.