Q.The standard electrode potential for Daniell cell is $1.1\ V$. Calculate the standard Gibbs energy for the reaction:
$Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$

Concept: Cell Representation & Nernst Equation — Gibbs energy is directly linked to cell potential via $\Delta G^\circ = -nFE^\circ$.

Step 1: Identify the number of electrons transferred ($n$).

$Zn \rightarrow Zn^{2+} + 2e^-$ and $Cu^{2+} + 2e^- \rightarrow Cu$, so $n = 2$.

Step 2: Use the standard Gibbs energy formula:

$\Delta G^\circ = -nFE^\circ_{\text{cell}}$

Step 3: Substitute values. Faraday constant $F = 96485\ \text{C mol}^{-1}$, $E^\circ_{\text{cell}} = 1.1\ \text{V}$. …

The key insight here is that electrochemical cells convert chemical energy into electrical work. The Nernst equation tells us how cell potential varies with concentration, but at standard conditions, the relationship between Gibbs energy and cell potential is beautifully simple: the maximum electrical work the cell can do equals the decrease in Gibbs energy.

For a spontaneous reaction like this one, $\Delta G^\circ$ must be negative, and the positive cell potential confirms that. The magnitude tells us how much useful work we can extract per mole of reaction.

1. Identify the number of electrons transferred.

   In the reaction $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$, zinc goes from oxidation state 0 to +2 (loses 2 electrons), and copper goes from +2 to 0 (gains 2 electrons). So $n = 2$ moles of electrons are transferred per mole of reaction.

2. Recall the fundamental relation.

   The standard Gibbs energy change is related to the standard cell potential by:

where $F$ is the Faraday constant ($96485\ \text{C mol}^{-1}$), $n$ is the number of electrons, and $E^\circ$ is the standard cell potential.

3. Plug in the values. $n = 2$, $F = 96485\ \text{C mol}^{-1}$, $E^\circ = 1.1\ \text{V}$.

4. Convert to kilojoules. …

Method: Gibbs Energy from Cell Potential (Nernst–Gibbs Relation)

This method uses the direct thermodynamic link between cell potential and Gibbs energy under standard conditions.

Key Formula

The standard Gibbs energy change $\Delta_r G^\circ$ is related to the standard cell potential $E^\circ_\text{cell}$ by:

Where:

• $n$ = number of moles of electrons transferred in the balanced reaction
• $F$ = Faraday constant = 96485 C mol⁻¹ (or 96500 C mol⁻¹ in many exam contexts)
• $E^\circ_\text{cell}$ = standard cell potential (in volts)

Step-by-Step Solution

Step 1: Identify $n$ (electrons transferred)

Write the half-reactions:

• Oxidation (anode): $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$
• Reduction (cathode): $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$

Each half-reaction involves 2 electrons. So:

Step 2: Note given data

• $E^\circ_\text{cell} = 1.1\ \text{V}$
• $F = 96485\ \text{C mol}^{-1}$ (use 96500 if exam specifies)

Step 3: Apply the formula

Step 4: Calculate

First compute: …

Here’s a breakdown of the common mistakes students make when solving this type of problem, along with how to avoid each one.

1. Forgetting the Sign Convention for $E^\circ_{\text{cell}}$

The Mistake:

Students often use the wrong sign for the standard cell potential, especially when they calculate it from half-reactions. They might subtract incorrectly or forget that the Daniell cell has a positive $E^\circ_{\text{cell}} = +1.1\ \text{V}$.

Why it happens:

Confusion between $E^\circ_{\text{cathode}}$ and $E^\circ_{\text{anode}}$ leads to sign errors.

How to avoid:

• For a spontaneous reaction (like Daniell cell), $E^\circ_{\text{cell}}$ is always positive.
• Use the formula:

• In the Daniell cell:
  • Cathode: $Cu^{2+}/Cu$ ($E^\circ = +0.34\ \text{V}$)
  • Anode: $Zn^{2+}/Zn$ ($E^\circ = -0.76\ \text{V}$)
  • So: $E^\circ_{\text{cell}} = 0.34 - (-0.76) = +1.10\ \text{V}$
• Always check: If the reaction is spontaneous, $E^\circ_{\text{cell}} > 0$.

2. Using the Wrong Number of Electrons ($n$)

The Mistake:

Students take $n = 1$ or $n = 3$ instead of the correct value from the balanced equation.

Why it happens:

They forget to balance the half-reactions or misread the stoichiometry.

How to avoid:

• Write the balanced half-reactions:
  • Oxidation: $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$
  • Reduction: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$
• Count the electrons transferred: Here, $n = 2$.
• Rule: $n$ is the number of moles of electrons that cancel when you add the half-reactions.
• Double-check: The overall reaction must have no free electrons.

3. Using the Wrong Value for Faraday’s Constant ($F$)

The Mistake:

Students use an inconsistent value for F (96485 vs 96500 C/mol) across a single solution, or forget to convert J to kJ at the end.

Why it happens:

Faraday’s constant has multiple common values, and unit confusion between Joules and kilojoules is common.

How to avoid:

• Use $F = 96485\ \text{C mol}^{-1}$ (matching the value used consistently elsewhere in this answer).
• Always check units:
  • $E^\circ$ is in Volts ($\text{J/C}$)
  • $n$ is unitless (moles of electrons)
  • $F$ is in $\text{C/mol}$
  • So $\Delta G^\circ = -nFE^\circ$ gives Joules (since $\text{V} \times \text{C} = \text{J}$)
• Convert to kJ by dividing by 1000 at the end.

4. Forgetting the Negative Sign in $\Delta G^\circ = -nFE^\circ$

The Mistake:

Students write $\Delta G^\circ = nFE^\circ$ (missing the minus sign) and get a positive value for a spontaneous reaction.

Why it happens:

They memorize the formula incorrectly or confuse it with the Nernst equation.

How to avoid:

• Memorize: $\Delta G^\circ = -nFE^\circ_{\text{cell}}$
• Logical check: For a spontaneous reaction ($E^\circ > 0$), $\Delta G^\circ$ must be negative.
• If your $\Delta G^\circ$ comes out positive, you likely missed the minus sign.

5. Not Converting Units Correctly (J to kJ)

The Mistake:

Students report $\Delta G^\circ$ in Joules when the answer is expected in kilojoules (or vice versa).

Why it happens:

They forget to divide by 1000 after calculation.

How to avoid:

• After calculating $\Delta G^\circ$ in Joules, always check if the problem asks for kJ.
• If so, divide by 1000. …