An aqueous solution containing `Na^(+),Sn^(2+),Cl^(-) " & " SO_(4)^(2-)` ions, all at unit concentration, is electrolysed between a silver anode and a platinum cathode. What changes occur at the electrodes when current is passed through the cell ? Given : `E_(Ag^(+)|Ag)^(0)=0.799 V`
`E_(Sn^(2+)|Sn)^(@)=-0.14V, E_(Cl_(2)|Cl^(-))^(@)=1.36 V, E_(S_(2)O_(8)^(2-)|SO_(4)^(2-))^(@)=2V, E_(Sn^(4+)|Sn^(2+))^(@)=0.13V`

Options

A. `Sin^(2+)` is reduced and `Cl^(-)` is oxidized.
B. Ag is oxidized and `Sn^(2+)` is reduced.
C. `Sn^(2+)` is reduced and `Sn^(2+)` is oxidized.
D. `H^(+)` is reduced and `Sn^(2+)` is oxidized.

Answer

C

Explanation

At anode, either Ag can get oxidised to `Ag^(+)` or `Sn^(2+)` to `Sn^(4+)` or `CI^(-)` to `CI_(2)` or `SO_(4)^(2-)` to `S_(2)O_(8)^(2-)`. Their respective oxidation potential values are `0.799 V,-0.13 V,-13.6V` and `-2V`. From these values, it is evident that `Sn^(2+)` would be oxidised first, followed by Ag at anode. At cathode, either `Na^(+)` can get reduced to `Na` or `Sn^(+)` to Sn or `H^(+)` to `H_(2)`. The reduction potential value of `Na^(+)` is highly negative while for `Sn^(2+)|Sn` is `-0.14V` and for `H^(+) +e^(-) rarr (1)/(2)H_(2)`
`(E_(H^(+)//H_(2)) =- 0.059 log.(1)/(10^(-7))) -0.413V`. Thus, `Sn^(2+)` will get reduced at cathode followed by `H^(+)`