Question

If you wanted to shift the potential of the lead-zinc cell upward,toward 0.70 V, which of the following actions would give thedesired result? Check all that apply. *Ignore the currentchecks.

Anyaction that makes [Zn2+] smaller than[Pb2+].
Any action that makes [Zn2+]larger than [Pb2+].
Adding some concentratedZn(C2H3O2)2 solution tothe Zn2+/Zn couple.
Anyaction that makes Q > 1.0.
Addingsome concentratedPb(C2H3O2)2 solution tothe Pb2+/Pb couple.
Anyaction that makes log Q positive.
Any action that makes log Qnegative.
Any action that makes Q <1.0.

Answer 1

To shift the potential of the lead-zinc cell upward toward 0.70 V, the following actions would give the desired result:

- Any action that makes [Zn2+] larger than [Pb2+].

- Adding some concentrated Zn(C2H3O2)2 solution to the Zn2+/Zn couple.

- Adding some concentrated Pb(C2H3O2)2 solution to the Pb2+/Pb couple.

- Any action that makes log Q negative.

- Any action that makes Q < 1.0.

To shift the potential of the lead-zinc cell upward toward 0.70 V, we need to consider the Nernst equation, which relates the cell potential (Ecell) to the concentrations of the ions involved in the cell reaction:

Ecell = E°cell - (RT/nF) * ln(Q)

Where:

- Ecell is the cell potential.

- E°cell is the standard cell potential.

- R is the gas constant.

- T is the temperature in Kelvin.

- n is the number of moles of electrons transferred in the cell reaction.

- F is the Faraday constant.

- Q is the reaction quotient.

In this case, we want to increase the cell potential (make it more positive), which means we want to make the logarithmic term (ln(Q)) more negative. To do that, we should consider the factors that affect Q:

1. Any action that makes [Zn2+] smaller than [Pb2+]:

- This would increase Q, making ln(Q) more positive, which is not what we want.

2. Any action that makes [Zn2+] larger than [Pb2+]:

- This would decrease Q, making ln(Q) more negative, which is what we want.

3. Adding some concentrated Zn(C2H3O2)2 solution to the Zn2+/Zn couple:

- This would increase the concentration of Zn2+, making Q smaller, which is what we want.

4. Any action that makes Q > 1.0:

- This would make ln(Q) more positive, which is not what we want.

5. Adding some concentrated Pb(C2H3O2)2 solution to the Pb2+/Pb couple:

- This would increase the concentration of Pb2+, making Q smaller, which is what we want.

6. Any action that makes log Q positive:

- This would make ln(Q) more positive, which is not what we want.

7. Any action that makes log Q negative:

- This would make ln(Q) more negative, which is what we want.

8. Any action that makes Q < 1.0:

- This would make ln(Q) more negative, which is what we want.

Answer 2

The actions that are likely to give the desired result are:

• Any action that makes [Zn2+] smaller than [Pb2+].
• Adding some concentrated Zn(C2H3O2)2 solution to the Zn2+/Zn couple.
• Any action that makes
  `\(\log Q\)` negative.
• Any action that makes
  `\(Q < 1.0\)`.

To shift the potential of the lead-zinc cell upward toward 0.70 V, the Nernst equation is a useful guide:

`\[E_{\text{cell}} = E^\circ_{\text{cell}} - (0.0592)/(n) \log Q\]`

Where:

• 
  `\(E_{\text{cell}}\)` is the cell potential,
• 
  `\(E^\circ_{\text{cell}}\)` is the standard cell potential,
• 
  `\(n\)` is the number of moles of electrons transferred,
• 
  `\(Q\)` is the reaction quotient.

Therefore, the actions that would give the desired result are:

• Any action that makes [Zn2+] smaller than [Pb2+].
• Adding some concentrated Zn(C2H3O2)2 solution to the Zn2+/Zn couple.
• Any action that makes
  `\(\log Q\)` negative.
• Any action that makes
  `\(Q < 1.0\)`.