1)
At anode oxidation occurs.
Here In(s) is being oxidised to In3+(aq)
So, the reaction at anode is:
In(s) -> In3+(aq) + 3e-
2)
At cathode reduction occurs.
Here Cd2+(aq) is being reduced to Cd(s)
So, the reaction at cathode is:
Cd2+(aq) + 2e- -> Cd(s)
3)
Balance the number of electrons to 6 in each case.
2 In(s) -> 2 In3+(aq) + 6e-
3 Cd2+(aq) + 6e- -> 3 Cd(s)
Combine both reaction to get net cell reaction which is”
2 In(s) + 3 Cd2+(aq) -> 2 In3+(aq) + 3 Cd(s)
Resources Check Answer Question 24 of 24 > Write the half-reactions as they occur at each...
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