I have no idea what I'm doing here, please help.
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.
I think this was reposted above with answers and I responded.
I'd be happy to help you understand the concept of galvanic cells and electrolytic cells, as well as how to determine the oxidation and reduction half reactions, overall cell reactions, and the effect on the mass of the cathode and anode.
A galvanic cell, also known as a voltaic cell, is an electrochemical cell that uses spontaneous redox reactions to generate electrical energy. On the other hand, an electrolytic cell is an electrochemical cell that requires an external electrical energy source to drive a non-spontaneous redox reaction.
a) Let's start by determining the oxidation and reduction half reactions in the galvanic cell you described:
At the anode (negative electrode):
The iron electrode (Fe(s)) in the solution of iron (II) sulfate (FeSO4(aq)) will undergo oxidation. The oxidation half reaction can be written as:
Fe(s) ⟶ Fe2+(aq) + 2e-
At the cathode (positive electrode):
The silver electrode (Ag(s)) in the silver nitrate solution will undergo reduction. The reduction half reaction can be written as:
Ag+(aq) + e- ⟶ Ag(s)
To determine the overall cell reaction, we need to balance the number of electrons lost in the oxidation half reaction and the number of electrons gained in the reduction half reaction. Multiplying the oxidation half reaction by 2 to balance the electrons:
2Fe(s) ⟶ 2Fe2+(aq) + 4e-
Finally, by adding the two half reactions together, we can obtain the overall cell reaction:
2Fe(s) + 2Ag+(aq) ⟶ 2Fe2+(aq) + 2Ag(s)
Simplifying the reaction, we can eliminate any spectator ions, which do not participate in the redox reactions:
Fe(s) + 2Ag+(aq) ⟶ Fe2+(aq) + 2Ag(s)
Now, regarding the mass of the cathode and anode:
During the galvanic cell operation, the anode (Fe) will be oxidized, losing mass as it forms Fe2+(aq). At the same time, the cathode (Ag) will be reduced, gaining mass as it forms Ag(s). Therefore, the mass of the cathode will increase, while the mass of the anode will decrease.
b) Now let's consider the electrolytic cell scenario:
In an electrolytic cell, the non-spontaneous redox reaction is driven by an external electrical energy source. The overall cell reaction will be the same as in the galvanic cell, but the direction of the electron flow will be reversed.
The oxidation half reaction at the anode remains the same:
2Fe(s) ⟶ 2Fe2+(aq) + 4e-
In contrast, the reduction half reaction at the cathode will be different since the objective is to plate Ag(s) onto the electrode:
Ag+(aq) + e- ⟶ Ag(s)
Therefore, the overall cell reaction for the electroplating of Ag will be:
2Fe(s) + 2Ag+(aq) ⟶ 2Fe2+(aq) + 2Ag(s)
Similar to the galvanic cell, the mass of the cathode will increase due to the deposition of Ag(s) onto the electrode. However, in the electrolytic cell, since the oxidation of Fe(s) doesn't occur, the mass of the anode will not decrease.
I hope this explanation helps you understand the concepts and processes involved in determining the half reactions, overall cell reactions, and how the mass of the electrodes is affected in both galvanic and electrolytic cells.