Question

write electrode reaction and cell reaction of a cell Al/Al3+/Cu2+/Cu.

Answer 1

Explanation: For a given cell $Al\mid Al^{3+}(aq.)\parallel Cu^{2+}(aq.)\mid Cu$, half-cell reactions can be written as:

Oxidation reaction: Whenever there is loss of electrons or increase in oxidation number, oxidation occurs.

$Al\rightarrow Al^{3+}+3e^-$             ..... (A)

Reduction reaction: Whenever there is gain of electrons or decrease in oxidation number, reduction occurs.

$Cu^{2+}(aq.)+2e^-\rightarrow Cu$     ......(B)

Cell reaction: In electrolysis it is known as overall reaction where total number of electrons remains same on both the sides of the reaction.

For a given cell, number of electrons on both the sides of the reaction are not same, hence we multiply equation A by 2 and equation B by 3.

$\text{Oxdation: }2Al\rightarrow 2Al^{3+}(aq.)+6e^-\\\text{Reduction: }3Cu^{2+}(aq.)+6e^-\rightarrow 3Cu\\\text{Net Cell reaction: }2Al+3Cu^{2+}(aq.)\rightarrow 2Al^{3+}(aq.)+3Cu$