The galvanic cell corresponding to the given redox reaction can be represented as :
`Zn|Zn_((aq))^(2+)||Ag_((aq))^(+) | Ag`
(i) Zn electrodes is negatively charged because at this electrode , Zn oxidizes to `Zn^(2+)` and the leaving electrons accumulate on this electrode .
(ii) Ions are the carriers of current in the cell.
(iii) The reaction taking place at Zn electrode can be represented as :
`Zn_((s)) to Zn_((aq))^(2+) + 2e^(-)`
And the reaction taking place at Zn electrode can be represented as :
`Ag_((aq))^(2+) + e^(-) to Ag_((s))`
(iv) In aqueous solutions, `CuCl_(2)` ionizes to give `Cu^(2+)` and `Cl^(-)` ions as :
`CuCl_(2 (aq)) to Cu_((aq))^(2+) + 2Cl_((aq))^(-)`
On electrolysis , either of `Cu^(2+)` ions or `H_(2)O` molecules can get reduced at the cathode . But the reduction potential of `Cu^(2+)` is more than that of `H_(2)O` molecules .
`Cu_((aq))^(2+) + 2 e^(-) to Cu_((aq)) , E^(@) = + 0.34V`
`H_(2)O_((l)) + 2e^(-) to H_(2 (g)) + 2OH^(-) , E^(@) = - 0.83V`
Hence , `Cu^(2+)` ions are reduced at the cathode and get deposited .
Similarly , at the anode , either of `Cl^(-)` or `H_(2)O`is oxidized . The oxidation potential of `H_(2)O` is higher than that of `Cl^(-)` .
`2Cl_((aq)) to Cl_(2(g)) + 2e^(-) , E^(@) = -1.36 V`
`2H_(2)O_((l)) to O_(2 (g)) + 4 H_((aq))^(+) + 4e^(-) , E^(@) = -1.23 V`
But oxidation of `H_(2)O` molecules occurs at a lower electrode potential than that of `Cl^(-)` ions because of over -voltage (extra voltage required to liberate gas ) . As a result , `Cl^(-)` ions are oxidized at the anode to liberate `Cl_(2)` gas .
