so i've worked on this some more, now I am asking if someone will plz check my work
2 half cells in a galvanic cell consist of one iron Fe(s) electorde 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 helf 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)
oxidation: Fe(s) -->Fe+2(aq) + 2e-
Reduction: Ag+(aq) + 2e- -->Ag(s)
the massof cathode will grow bigger because it's gaining electrons
b)Ag(s) + Fe(aq) + energy -->Ag+(aq) + Fe(s)
Oxidation: Ag(s) -->Ag+2(aq) + 2e-
Reduction: Fe2+(aq) + 2e- -->Fe(s)
mass of cathode iwll grow bigger because its gaining electrons
a) Your overall cell reaction is correct:
Fe(s) + 2Ag+(aq) -> Fe2+(aq) + 2Ag(s)
However, your individual half-reactions are not correct. The correct oxidation half-reaction should be:
Fe(s) -> Fe2+(aq) + 2e-
The correct reduction half-reaction should be:
Ag+(aq) + e- -> Ag(s)
b) In an electrolytic cell, the direction of electron flow is reversed compared to a galvanic cell. The correct overall cell reaction would be:
Ag+(aq) + Fe2+(aq) + energy -> Ag(s) + Fe(s)
The correct oxidation half-reaction would be:
Ag(s) -> Ag+(aq) + e-
The correct reduction half-reaction would be:
Fe2+(aq) + 2e- -> Fe(s)
In an electrolytic cell, the mass of the cathode will gain mass as it is the site of reduction and is gaining electrons. The mass of the anode will decrease as it is the site of oxidation and is losing mass.
To check your work, let's go through it step by step.
a) In a galvanic cell, the oxidation half-reaction occurs at the anode (negative electrode), while the reduction half-reaction occurs at the cathode (positive electrode).
The oxidation half-reaction is: Fe(s) -> Fe2+(aq) + 2e- (You have correctly identified this)
The reduction half-reaction is: Ag+(aq) + e- -> Ag(s) (You missed the charge on silver in the reduction half-reaction)
To balance the overall cell reaction, multiply the oxidation half-reaction by 2 and combine it with the reduction half-reaction:
2Fe(s) + 2Ag+(aq) -> 2Fe2+(aq) + 2Ag(s)
Eliminate spectator ions (those that appear on both sides of the equation) to obtain the overall cell reaction:
2Fe(s) + 2Ag+(aq) -> 2Fe2+(aq) + 2Ag(s)
During the operation of the cell, iron will lose mass at the anode (Fe(s) electrode) as it is oxidized to Fe2+(aq), while silver will gain mass at the cathode (Ag(s) electrode) as it is reduced from Ag+(aq) to Ag(s).
b) In an electrolytic cell, the process is reversed. The cathode becomes the negative electrode, where reduction occurs, and the anode becomes the positive electrode, where oxidation occurs.
The oxidation half-reaction is: Ag(s) -> Ag+(aq) + e- (You missed the charge on silver in the oxidation half-reaction)
The reduction half-reaction is: Fe2+(aq) + 2e- -> Fe(s) (You have correctly identified this)
To balance the overall cell reaction, you need to combine the oxidation and reduction half-reactions:
Ag(s) + Fe2+(aq) -> Ag+(aq) + Fe(s)
During the operation of the electrolytic cell, the cathode (Ag(s) electrode) will gain mass as it is reduced from Ag+(aq) to Ag(s), and the anode (Fe(s) electrode) will lose mass as it is oxidized from Fe2+(aq) to Fe(s).
Overall, you have correctly identified the half-reactions but missed the charges on silver in both cases. Make sure to always include the charges to balance the reactions correctly.
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-
OK
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)
OK
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.