1) Assume that the reference half-cell is changed to a standard mercury-mercury (II) half-cell.
a) What would be the reduction potential of a standard chlorine half-cell.
c) What would be the cell potential of a standard chlorine-nickel cell.
2) For each of the following standard cells, write the cell notation, label electrodes, and determine the cell potential.
a) tin(IV) zinc standard cell.
If someone could explain the questions/help me with them so I can do the remaining work it would be greatly appreciated. Thanks!
Sure! I can help you with these questions.
1) a) To determine the reduction potential of a standard chlorine half-cell, we first need to know the reduction potential of the standard mercury-mercury (II) half-cell. The standard reduction potential of the standard mercury-mercury (II) half-cell is 0.85 V.
To find the reduction potential of the chlorine half-cell, known as E°(Cl2/Cl-), we use the Nernst equation:
E°(cell) = E°(cathode) - E°(anode)
Since the standard hydrogen electrode (SHE) is commonly used as the reference electrode, we can find E°(cathode) as the reduction potential of the SHE, which is 0 V.
Therefore, E°(cell) = 0.85 V - 0.00 V = 0.85 V
So, the reduction potential of the standard chlorine half-cell is 0.85 V.
b) To find the cell potential of a standard chlorine-nickel cell, we need to know the reduction potentials of both half-cells: the standard chlorine half-cell and the standard nickel half-cell.
The standard reduction potential of the standard nickel half-cell, known as E°(Ni2+/Ni), is -0.25 V.
Using the same Nernst equation as before:
E°(cell) = E°(cathode) - E°(anode)
E°(cell) = E°(Ni2+/Ni) - E°(Cl2/Cl-)
E°(cell) = -0.25 V - 0.85 V = -1.10 V
So, the cell potential of a standard chlorine-nickel cell is -1.10 V.
2) For the tin(IV) zinc standard cell, we need to write the cell notation and determine the cell potential.
The cell notation consists of two half-cell reactions separated by a double vertical line. The anode (oxidation half-reaction) is written on the left side of the double line, and the cathode (reduction half-reaction) is written on the right side.
The oxidation half-reaction is the tin(IV) electrode releasing 2 electrons to become Sn(II):
Sn(IV) → Sn(II) + 2e-
The reduction half-reaction is the zinc electrode receiving the 2 electrons to become Zn(II):
Zn(II) + 2e- → Zn
The cell notation for the tin(IV) zinc standard cell is:
Sn(IV) | Sn(II) || Zn(II) | Zn
To determine the cell potential, we need to look up the reduction potentials of the two half-cells involved.
The standard reduction potential for tin(IV) to tin(II) half-cell, E°(Sn4+/Sn2+), is +0.15 V.
The standard reduction potential for zinc to zinc(II) half-cell, E°(Zn2+/Zn), is -0.76 V.
Using the Nernst equation, we can calculate the cell potential:
E°(cell) = E°(cathode) - E°(anode)
E°(cell) = E°(Zn2+/Zn) - E°(Sn4+/Sn2+)
E°(cell) = -0.76 V - (+0.15 V) = -0.91 V
Therefore, the cell potential of the tin(IV) zinc standard cell is -0.91 V.
I hope this explanation helps! Let me know if you have any further questions.