Question

Calculate emf of the following cell at 298 K :
\( \text{Zn (s)| Zn}^{2+}\text{(aq) (0.1 M)|| Ag}^+\text{ (aq) (0.01 M)|Ag (s)} \)
(Given : \( E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \text{ V}, E^\circ_{\text{Ag}^{+}/\text{Ag}} = +0.80 \text{ V, [log 10 = 1]} \))

Hint:

Always square the concentration of \( \text{Ag}^+ \) in the log term because of the stoichiometric coefficient 2.
Remember \( \log(10^x) = x \).
A positive \( E_{\text{cell}} \) indicates a spontaneous reaction.

Answer 1

Concept:

• The emf of a cell under non-standard conditions is calculated using the Nernst Equation.

• The overall cell reaction must be determined to identify the number of electrons \( (n) \) transferred.
Step 1: Identify the half-reactions and determine n
Anode (Oxidation): \( \text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2e^- \)
Cathode (Reduction): \( 2\text{Ag}^+\text{(aq)} + 2e^- \rightarrow 2\text{Ag(s)} \)
Overall reaction: \( \text{Zn(s)} + 2\text{Ag}^+\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{Ag(s)} \)
Number of electrons transferred, \( n = 2 \).

Step 2: Calculate the standard cell potential \( (E^\circ_{\text{cell}}) \)
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
\[ E^\circ_{\text{cell}} = 0.80 \text{ V} - (-0.76 \text{ V}) = 1.56 \text{ V} \]

Step 3: Apply the Nernst Equation
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Ag}^+]^2} \right) \]
Substituting the given values:
\[ E_{\text{cell}} = 1.56 - \frac{0.059}{2} \log \left( \frac{0.1}{(0.01)^2} \right) \]
\[ E_{\text{cell}} = 1.56 - 0.0295 \log \left( \frac{10^{-1}}{10^{-4}} \right) \]
\[ E_{\text{cell}} = 1.56 - 0.0295 \log(10^3) = 1.56 - (0.0295 \times 3) \]
\[ E_{\text{cell}} = 1.56 - 0.0885 = 1.4715 \text{ V} \] Final Answer: The emf of the cell is 1.4715 V.