| Half reaction | Standard reduction potential (V) |
| \[N{{i}^{2+}}(aq)+2{{e}^{-}}\to Ni(s)\] | - 0.236 |
| \[S{{n}^{2+}}(aq)+2{{e}^{-}}\to Sn(s)\] | - 0.141 |
| \[B{{r}_{2}}(aq)+2{{e}^{-}}\to 2B{{r}^{-}}(aq)\] | 1.077 |
| \[C{{l}_{2}}(aq)+2{{e}^{-}}\to 2C{{l}^{-}}(aq)\] | 1.360 |
A) Ni (s) at cathode, \[C{{l}_{2}}\](aq) at anode
B) Ni (s) at cathode, \[B{{r}_{2}}\] (aq) at anode
C) Sn (s) at cathode, \[B{{r}_{2}}\] (aq) at anode
D) Sn (s) at cathode, \[C{{l}_{2}}\] (aq) at anode
Correct Answer: C
Solution :
Metal with higher reduction potential (here Sn) will be first deposited at cathode and anion with higher oxidation potential (here\[B{{r}^{-}}\]) will be oxidised at anode.
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