Two half cells in a galvanic cell consist of one iron electroide in a solution of iron sulphate and a silver electrode in a silver nitrate solution

a) Assume that the cell is operating as a galvanic cell. State the oxidation half reaction, the reduction half reaction and the overall cell reaction. Describe what willhappen 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 anelectrolytic cell

Fe^+2 + 2e ==> Fe Eo = ??

Ag^+1 + e ==> Ag Eo = ??

Look up Eo values for each, reverse the one with the larger negative value and add the two half cells. You should obtain a + voltage.
a) oxidation occurs at the anode and reduction at the cathode; multiply the Ag reaction by 2 and add the two half cells (one oxidation and one reduction) to obtain the cell reaction.
b)If you get through a you should be able to do b on your own.

a) In a galvanic cell, the oxidation half-reaction occurs at the anode (negative electrode) and involves the loss of electrons, while the reduction half-reaction occurs at the cathode (positive electrode) and involves the gain of electrons.

For the given cell, the half-reactions can be determined as follows:
1. Oxidation half-reaction:
Iron electrode (anode): Fe(s) -> Fe^2+(aq) + 2e^-
In this half-reaction, solid iron (Fe) loses two electrons to form aqueous iron ions (Fe^2+).

2. Reduction half-reaction:
Silver electrode (cathode): Ag^+(aq) + e^- -> Ag(s)
In this half-reaction, silver ions (Ag^+) from the solution gain one electron to form solid silver (Ag).

To find the overall cell reaction, we need to multiply the oxidation half-reaction by two to balance the electrons:
2Fe(s) -> 2Fe^2+(aq) + 4e^-

Now we can combine the half-reactions to obtain the overall cell reaction:
2Fe(s) + Ag^+(aq) -> 2Fe^2+(aq) + 2e^- + Ag(s)
The overall cell reaction is the sum of the reduction and oxidation half-reactions.

Regarding the mass of the cathode and the anode, during the operation of the galvanic cell, the mass of the cathode (silver electrode) will increase as silver ions from the solution are reduced and deposited onto the electrode surface. At the same time, the mass of the anode (iron electrode) will decrease as iron atoms are oxidized and dissolve into the solution as Fe^2+ ions. This results in the transfer of mass from the anode to the cathode.

b) In an electrolytic cell, an external electrical source is used to drive a non-spontaneous reaction. The cathode becomes the positive electrode, and the anode becomes the negative electrode.

Assuming the given cell is an electrolytic cell, the half-reactions will remain the same, but the electron flow will be reversed. The reduction half-reaction will occur at the cathode (positive electrode), and the oxidation half-reaction will occur at the anode (negative electrode). So, the reactions would be as follows:

1. Reduction half-reaction at the cathode:
Ag^+(aq) + e^- -> Ag(s)

2. Oxidation half-reaction at the anode:
Fe(s) -> Fe^2+(aq) + 2e^-

The overall cell reaction will be the same as in the galvanic cell:

2Fe(s) + Ag^+(aq) -> 2Fe^2+(aq) + 2e^- + Ag(s)

In an electrolytic cell, the flow of electrons from the external source will cause the opposite result in terms of mass changes at the electrodes compared to a galvanic cell. The cathode (positive electrode, silver) will lose mass as silver atoms dissolve and enter the solution as Ag^+ ions. Meanwhile, the anode (negative electrode, iron) will gain mass as iron ions in the solution are reduced and deposited onto the electrode surface as solid iron.