Little confusing!! I need help PLZ!!!
Two 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. Assume the cell is operating as a galvanic cell, state the overall and half-cell reactions.
Describe what will happen to the mass of the cathode while the cell is operating.
b. Repeat part a), assuming that the cell is operating as an electrolytic cell
The iron half cell is
Fe(s) ==> Fe^2+(aq) + 2e
The Ag half cell is
Ag^+(aq) + e ==> Ag(s)
Overall is
Fe(s) + 2Ag^+(aq) ==> Ag(s) + Fe^2+(aq)
I have written (aq) just to show Ag or Fe is in solution; the usual procedure is to place the concn in M there.
The anode is Fe; the cathode is Ag. Ag plates out on the cathode; (Fe goes into solution at the anode)
Part 2 is the reverse of all of this in an electrolytic cell.
a. In a galvanic cell, the overall reaction can be determined by combining the half-cell reactions.
The half-cell reaction at the anode (oxidation half-reaction) involves the iron electrode:
Fe(s) -> Fe2+(aq) + 2e-
The half-cell reaction at the cathode (reduction half-reaction) involves the silver electrode:
Ag+(aq) + e- -> Ag(s)
Combining these two half-cell reactions gives the overall reaction of the galvanic cell:
Fe(s) + Ag+(aq) -> Fe2+(aq) + Ag(s)
During the operation of the galvanic cell, the iron electrode will dissolve, resulting in a decrease in its mass. This is because the iron electrode is the anode, where oxidation occurs and the iron atoms lose electrons to become Fe2+ ions in solution.
b. In an electrolytic cell, the overall reaction is driven by an external source of electrical energy. Therefore, the overall reaction will be the reverse of the galvanic cell reaction.
The half-cell reaction at the anode (oxidation half-reaction) remains the same as in the galvanic cell:
Fe2+(aq) + 2e- -> Fe(s)
The half-cell reaction at the cathode (reduction half-reaction) also remains the same:
Ag(s) -> Ag+(aq) + e-
However, since an external energy source is driving the reaction, the overall reaction will be the reverse of the galvanic cell reaction:
Fe2+(aq) + Ag(s) -> Fe(s) + Ag+(aq)
In this case, the iron electrode will gain mass during the operation of the electrolytic cell. This is because the iron electrode is now the cathode, where reduction occurs, and the Fe2+ ions in solution gain electrons to become neutral iron atoms, depositing on the electrode and increasing its mass.
a) To determine the overall and half-cell reactions for the given galvanic cell, we need to understand the redox reactions happening at each electrode.
At the anode (oxidation half-cell), iron atoms are being oxidized to Fe2+ ions. This can be represented by the following half-cell reaction:
Fe(s) → Fe2+(aq) + 2e-
At the cathode (reduction half-cell), silver ions (Ag+) accept electrons and get reduced to silver atoms that deposit on the electrode. This can be represented by the following half-cell reaction:
Ag+(aq) + e- → Ag(s)
To obtain the overall reaction, we need to balance the number of electrons transferred in the half-cell reactions. In the oxidation half-cell, 2 electrons are released, while in the reduction half-cell, only 1 electron is consumed. So, we need to multiply the reduction half-cell reaction by 2:
2(Ag+(aq) + e-) → 2Ag(s)
Adding the two half-cell reactions together, we get the overall reaction:
Fe(s) + 2Ag+(aq) → Fe2+(aq) + 2Ag(s)
Now, regarding the second part of the question:
When operating as a galvanic cell, the mass of the cathode (silver electrode) will increase over time. This happens because silver ions from the silver nitrate solution are reduced and deposited as solid silver atoms onto the cathode. So, the silver electrode will gain mass as silver ions are converted into silver metal.
b) In an electrolytic cell, the process is reversed. Instead of a spontaneous redox reaction, an external source of electrical energy is used to force a non-spontaneous redox reaction to occur. This means that the oxidation and reduction processes are reversed compared to a galvanic cell.
In this case, at the anode (positive electrode), silver metal (Ag(s)) will undergo oxidation and lose electrons to form silver ions (Ag+):
Ag(s) → Ag+(aq) + e-
At the cathode (negative electrode), iron ions (Fe2+) from the iron (II) sulphate solution will accept electrons and get reduced to iron metal (Fe(s)):
Fe2+(aq) + 2e- → Fe(s)
The overall reaction for this electrolytic cell can be obtained by balancing the electron transfer in the half-cell reactions:
2Ag(s) → 2Ag+(aq) + 2e-
Fe2+(aq) + 2e- → Fe(s)
Adding these half-cell reactions gives the overall reaction for the electrolytic cell:
2Ag(s) + Fe2+(aq) → 2Ag+(aq) + Fe(s)
In an electrolytic cell, the mass of the cathode would increase as iron ions are reduced and deposited as solid iron onto the cathode.