I don't think you have it yet but I have made some comments below (in bold).
2 half cells in a galvanic cell consist of one iron Fe(s) electrode in a solution of iron (II) sulphate FeSO4(aq)and a silver Ag(s) electrode in a silver nitrate solution,
a)state the oxidation half reaction, the reduction half reaction and the overall cell reaction. remember to eliminiate any spectator ions. describe what will happen to the mass of the cathode and the mass of the anode while the cell is operating.
b)repeat part a, assuming that the cell is operating as an electrolytic cell.
a)Fe(s) -->Fe+2(aq) + Ag(s)
Is this the overall reaction? If so, it should be
Fe(s) + 2Ag^+(aq) ==>2Ag(s) + Fe^+2(aq)
oxidation: Fe(s) -->Fe+2(aq) + 2e-
Reduction: Ag+(aq) + 2e- -->Ag(s)
Ag^+(aq) + e ==> Ag(s)
the mass of cathode will grow bigger because it's gaining electrons
You are correct that it will grow bigger (I prefer to say that the Ag electode gains mass) BUT not because it gains electrons. Considering that an electron has a mass of only about 10^-28 g. The silver electrode gains mass because silver is depositing on it. When the cell is operating, the Fe electrode is going into solution (look at the equation you wrote for the Fe electrode) and the silver ions are being deposited on the silver electrode (again, look at the half equations you wrote to see that).
b)Ag(s) + Fe(aq) + energy -->Ag+(aq) + Fe(s)
electrons are not balanced, Fe(aq) shown with no charge
Oxidation: Ag(s) -->Ag+2(aq) + 2e-
Ag^+ and not Ag^+2; therefore 1e- instead of 2e-.
Reduction: Fe2+(aq) + 2e- -->Fe(s)
mass of cathode iwll grow bigger because its gaining electrons.
The cathode (the Fe electrode) will gain in mass because Fe^+2 + 2e ==> Fe is occurring and Fe is being deposited on the Fe electrode.
Let me know if anything is not clear.
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