`Cr_(2)O_(7)^(2-)+I^(-)toI_(2)+Cr^(3+)`
`E_(cell)^(@)=0.79V,E_(Cr_(2)O_(7)^(2-))^(@)=1.33V,E_(I_(2))^(o)` is

Options

A. `-0.10 V`
B. `+0.18 V`
C. `-0.54 V`
D. 0.54 V

Answer

D

Explanation

`I^(-)` get oxidised to `I_(2)` hence will form anode and `Cr_(2)O_(7)^(2-)` reduced to `Cr^(3+)` hence will form cathode.
`E_("cell")^(@) = E_("cathode")^(@) - E_("anode")^(@) : E_("cell")^(@) = E_(Cr_(2)O_(7)^(2-))^(@) - E_(I_(2))^(@)`
`0.79 = 1.33 - E_(I_(2))^(@) , E_(l_(2))^(@) = 1.33 - 0.79, E_(l_(2))^(@) = 0.54 V`