Q.In the button cells widely used in watches and other devices the following reaction takes place:
$Zn(s) + Ag_2O(s) + H_2O(l) \rightarrow Zn^{2+}(aq) + 2Ag(s) + 2OH^-(aq)$
Determine $\Delta_rG^\circ$ and $E^\circ$ for the reaction.

The key idea is to use the standard reduction potentials to find the cell potential, then relate it to Gibbs free energy via $\Delta_r G^\circ = -nFE^\circ$.

Step 1: Identify half-reactions and look up standard potentials.

Anode (oxidation): $\text{Zn(s)} \rightarrow \text{Zn}^{2+}(\text{aq}) + 2e^-$, $E^\circ_{\text{ox}} = +0.76\ \text{V}$ (since $E^\circ_{\text{red}}(\text{Zn}^{2+}/\text{Zn}) = -0.76\ \text{V}$).

Cathode (reduction): $\text{Ag}_2\text{O(s)} + \text{H}_2\text{O(l)} + 2e^- \rightarrow 2\text{Ag(s)} + 2\text{OH}^-(\text{aq})$, $E^\circ_{\text{red}} = +0.344\ \text{V}$.

Step 2: Calculate standard cell potential.

$E^\circ = E^\circ_{\text{cathode}} + E^\circ_{\text{anode}} = 0.344\ \text{V} + 0.76\ \text{V} = 1.104\ \text{V}$.

Step 3: Determine $n$ and compute $\Delta_r G^\circ$. …

This is a classic electrochemistry problem from a button cell — the kind used in watches, hearing aids, and small electronics. The reaction given is the overall cell reaction, and we need to find the standard Gibbs free energy change ($\Delta_rG^\circ$) and the standard cell potential ($E^\circ$).

The key idea: $E^\circ$ and $\Delta_rG^\circ$ are linked by $\Delta_rG^\circ = -nFE^\circ$, where $n$ is the number of moles of electrons transferred and $F$ is Faraday's constant ($96485\ \text{C mol}^{-1}$). So if we can find $E^\circ$ from standard reduction potentials, we can compute $\Delta_rG^\circ$.

Let’s break it down.

1. Identify the half-reactions. The overall reaction is:

Zinc is being oxidised: $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$

Silver oxide is being reduced. In basic medium, $Ag_2O$ reduces to $Ag$ with water and electrons:

Notice both half-reactions involve 2 electrons — so $n = 2$.

2. Look up standard reduction potentials.

   From standard tables (at $25^\circ C$, 1 M, 1 atm):

   • $Zn^{2+}(aq) + 2e^- \rightarrow Zn(s)$: $E^\circ = -0.76\ \text{V}$
   • $Ag_2O(s) + H_2O(l) + 2e^- \rightarrow 2Ag(s) + 2OH^-(aq)$: $E^\circ = +0.344\ \text{V}$
   Watch out

   A common mistake: using the reduction potential of $Ag^+$ instead of $Ag_2O$. The problem gives $Ag_2O$, not $Ag^+$, so use the correct value. Also, remember that the zinc half-reaction is written as a reduction — we will reverse it for oxidation.

3. Calculate $E^\circ_{cell}$.

   The standard cell potential is:

Here, reduction occurs at the cathode ($Ag_2O$), oxidation at the anode ($Zn$). So:

$E^\circ_{cell} = (+0.344\ \text{V}) - (-0.76\ \text{V}) = 1.104\ \text{V}$ …

Method: Standard Gibbs Energy from Standard Cell Potential

We use the Nernst–Gibbs relation connecting standard cell potential ($E^\circ$) and standard Gibbs energy change ($\Delta_r G^\circ$):

Where:

• $n$ = number of moles of electrons transferred
• $F$ = Faraday constant ($96485\ \text{C mol}^{-1}$)
• $E^\circ$ = standard cell potential (in volts)

Step 1: Identify the half-reactions and find $n$

Oxidation (anode):

Reduction (cathode):

Number of electrons transferred: $n = 2$

Step 2: Calculate $E^\circ$ for the cell

$E^\circ = \mathbf{+1.104\ V}$

Step 3: Calculate $\Delta_r G^\circ$

$\Delta_r G^\circ = -n F E^\circ$ …

Common Mistakes & How to Avoid Them

Mistake 1: Writing the Wrong Cell Representation

The error: Students often try to force the reaction into a standard Daniell cell format (e.g., Zn | Zn²⁺ || Ag⁺ | Ag), ignoring that Ag₂O and OH⁻ are involved.

Why it happens: The reaction includes solids (Ag₂O, Ag) and aqueous ions (Zn²⁺, OH⁻), but no free Ag⁺ ions. This confuses students who memorise only the simplest cell setups.

How to avoid:

• Identify the two half-reactions first — never jump to the cell diagram.
• For this reaction:
  • Anode (oxidation): $\text{Zn(s)} \rightarrow \text{Zn}^{2+}(\text{aq}) + 2e^-$
  • Cathode (reduction): $\text{Ag}_2\text{O(s)} + \text{H}_2\text{O(l)} + 2e^- \rightarrow 2\text{Ag(s)} + 2\text{OH}^-(\text{aq})$
• Only then write the cell notation: $\text{Zn(s)} \mid \text{Zn}^{2+}(\text{aq}) \parallel \text{OH}^-(\text{aq}) \mid \text{Ag}_2\text{O(s)}, \text{Ag(s)}$

Mistake 2: Using the Wrong $E^\circ$ Values from the Table

The error: Students pick $E^\circ_{\text{Ag}^+/\text{Ag}} = +0.80\ \text{V}$ instead of the correct value for the Ag₂O/Ag couple in basic medium.

Why it happens: Standard reduction potential tables list many silver couples. The most common one (Ag⁺/Ag) is not relevant here because the reaction involves Ag₂O and OH⁻, not free Ag⁺.

How to avoid:

• Check the species in your half-reaction. Here, the reduction is: $\text{Ag}_2\text{O(s)} + \text{H}_2\text{O(l)} + 2e^- \rightarrow 2\text{Ag(s)} + 2\text{OH}^-(\text{aq})$
• The correct standard potential is: $E^\circ_{\text{Ag}_2\text{O}/\text{Ag}} = +0.344\ \text{V}$
• Memorise key couples for common battery reactions, or always verify from the given data in the exam.

Mistake 3: Sign Errors When Calculating $E^\circ_{\text{cell}}$

The error: Students subtract in the wrong order or forget that the anode potential must be reversed.

Why it happens: The formula $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ is misapplied when students treat both as reduction potentials but swap the roles.

How to avoid:

• Always use reduction potentials for both half-cells.
• For this reaction:
  • Cathode (reduction): $E^\circ = +0.344\ \text{V}$
  • Anode (oxidation): reverse of Zn²⁺/Zn reduction. The reduction potential for Zn²⁺/Zn is $-0.76\ \text{V}$.
• Then: $E^\circ_{\text{cell}} = (+0.344) - (-0.76) = +1.104\ \text{V}$
• Check sign: A positive $E^\circ_{\text{cell}}$ confirms a spontaneous reaction — if you get negative, you likely swapped the half-cells.

Mistake 4: Forgetting $n$ in the $\Delta_r G^\circ$ Calculation

The error: Students use $n = 1$ or $n = 4$ instead of the correct number of electrons transferred.

Why it happens: The overall reaction has multiple electrons, but students miscount because they don't balance the half-reactions properly.

How to avoid:

• Balance each half-reaction and note the electron count.
  • Anode: $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$ → 2 electrons
  • Cathode: $\text{Ag}_2\text{O} + \text{H}_2\text{O} + 2e^- \rightarrow 2\text{Ag} + 2\text{OH}^-$ → 2 electrons
• They match, so $n = 2$.
• Always confirm that electrons cancel in the overall reaction.

Mistake 5: Rounding $E^\circ_{\text{cell}}$ Too Early, Which Skews $\Delta_r G^\circ$ …