Question 2(a)


In the reaction:

Sn(s) + CuSO4(aq) = Cu(s) + SnSO4(aq)

Sn(s) is oxidised to Sn2+(aq)
Cu2+(aq) is reduced to Cu(s).

Writing the half-cell reactions as reductions:

Cu2+(aq) + SO42-(aq) + 2e = Cu(s) + SO42-(aq)Sn2+(aq) + SO42-(aq) + 2e = Sn(s) + SO42-(aq)

The SO42-(aq) ions are unaltered in the reaction ("spectator ions") hence they may be ignored and we can write:

Cu2+(aq) + 2e = Cu(s)Sn2+(aq) + 2e = Sn(s)

This implies the two half-cells of interest are:

Cu2+(aq) | Cu(s)Sn2+(aq) | Sn(s)

Cu2+(aq) is reduced in the over-all reaction. Hence the Cu2+(aq)/Cu(s) must form the right-hand half-cell, in order that Cu2+(aq) is reduced in the formal cell reaction. Therefore the cell with the formal reaction:

Sn(s) + CuSO4(aq) = Cu(s) + SnSO4(aq)

is

Sn(s) | Sn2+(aq) || Cu2+(aq) | Cu(s)

Carry on checking the answer to Q2(a).
Return to the question.