Question

Standard electrode potentials for a few half-cells are mentioned below: 

• A. \( \text{Zn} | \text{Zn}^{2+} (1M) || \text{Ag}^+ (1M) | \text{Ag} \)
• B. \( \text{Zn} | \text{Zn}^{2+} (1M) || \text{Mg}^{2+} (1M) | \text{Mg} \)
• C. \( \text{Ag} | \text{Ag}^+ (1M) || \text{Mg}^{2+} (1M) | \text{Mg} \)
• D. \( \text{Cu} | \text{Cu}^{2+} (1M) || \text{Ag}^+ (1M) | \text{Ag} \)

Hint:

To determine the cell with the most negative \( \Delta G^\circ \), consider the cell with the largest difference in standard electrode potentials, where the anode has the most negative \( E^\circ \) and the cathode has the most positive \( E^\circ \).

Answer 1

The Correct Option is A

To solve this problem, we need to calculate the standard cell potentials \( E^0_{\text{cell}} \) of the given galvanic cells. The standard cell potential is determined using the formula: 

\(E^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}}\)

We will use the provided standard electrode potentials:

Half-cell	Standard Electrode Potential (V)
\(\text{Cu}^{2+} | \text{Cu}\)	+0.34
\(\text{Zn}^{2+} | \text{Zn}\)	-0.76
\(\text{Ag}^{+} | \text{Ag}\)	+0.80
\(\text{Mg}^{2+} | \text{Mg}\)	-2.37

1. Cell: \( \text{Zn} | \text{Zn}^{2+} (1M) || \text{Ag}^+ (1M) | \text{Ag} \)

Here, \(\text{Ag}^+\) is the cathode and \(\text{Zn}\) is the anode.

\(E^0_{\text{cell}} = E^0_{\text{Ag}^+/\text{Ag}} - E^0_{\text{Zn}^{2+}/\text{Zn}}\)

\(E^0_{\text{cell}} = (+0.80) - (-0.76) = +1.56 \, \text{V}\)

1. Cell: \( \text{Zn} | \text{Zn}^{2+} (1M) || \text{Mg}^{2+} (1M) | \text{Mg} \)

Here, \(\text{Mg}^{2+}\) is the cathode and \(\text{Zn}\) is the anode.

\(E^0_{\text{cell}} = E^0_{\text{Mg}^{2+}/\text{Mg}} - E^0_{\text{Zn}^{2+}/\text{Zn}}\)

\(E^0_{\text{cell}} = (-2.37) - (-0.76) = -1.61 \, \text{V}\)

1. Cell: \( \text{Ag} | \text{Ag}^+ (1M) || \text{Mg}^{2+} (1M) | \text{Mg} \)

Here, \(\text{Mg}^{2+}\) is the cathode and \(\text{Ag}\) is the anode.

\(E^0_{\text{cell}} = E^0_{\text{Mg}^{2+}/\text{Mg}} - E^0_{\text{Ag}^+/\text{Ag}}\)

\(E^0_{\text{cell}} = (-2.37) - (+0.80) = -3.17 \, \text{V}\)

1. Cell: \( \text{Cu} | \text{Cu}^{2+} (1M) || \text{Ag}^+ (1M) | \text{Ag} \)

Here, \(\text{Ag}^+\) is the cathode and \(\text{Cu}\) is the anode.

\(E^0_{\text{cell}} = E^0_{\text{Ag}^+/\text{Ag}} - E^0_{\text{Cu}^{2+}/\text{Cu}}\)

\(E^0_{\text{cell}} = (+0.80) - (+0.34) = +0.46 \, \text{V}\)

The cell with the highest positive standard cell potential is \( \text{Zn} | \text{Zn}^{2+} (1M) || \text{Ag}^+ (1M) | \text{Ag} \) with \(+1.56 \, \text{V}\).

Therefore, the correct answer is:

\( \text{Zn} | \text{Zn}^{2+} (1M) || \text{Ag}^+ (1M) | \text{Ag} \)

Answer 2

Step 1: Understand the cell notation format

A galvanic cell (electrochemical cell) is represented in the notation: \[ \text{Anode} \,|\, \text{Anode Solution (concentration)} \,||\, \text{Cathode Solution (concentration)} \,|\, \text{Cathode} \] where oxidation occurs at the anode and reduction occurs at the cathode.

Step 2: Identify electrode potentials

From the given table, the standard electrode potentials are:

• \( \text{Zn}^{2+}/\text{Zn} = -0.76\,\text{V} \)
• \( \text{Ag}^{+}/\text{Ag} = +0.80\,\text{V} \)

Step 3: Determine which is oxidized and which is reduced

- Lower (more negative) standard electrode potential: Zn is more likely to lose electrons (get oxidized).
- Higher (more positive) standard electrode potential: Ag⁺ is more likely to gain electrons (get reduced).

Step 4: Assign anode and cathode

• Anode (oxidation): Zn → Zn²⁺ + 2e⁻
• Cathode (reduction): Ag⁺ + e⁻ → Ag

Step 5: Write the cell notation

Following the convention (Anode | Anode solution || Cathode solution | Cathode), we get: \[ \text{Zn} \,|\, \text{Zn}^{2+} (1\,\text{M}) \,||\, \text{Ag}^{+} (1\,\text{M}) \,|\, \text{Ag} \]

Final Answer:
\( \boxed{\text{Zn} \,|\, \text{Zn}^{2+} (1\,\text{M}) \,||\, \text{Ag}^{+} (1\,\text{M}) \,|\, \text{Ag}} \)