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Q2.5(iii):
Write the Nernst equation and emf of the following cell at 298 K: \(Sn(s)|Sn^{2+}(0.050 M)||H^{+}(0.020 M)|H_{2}(g) (1 bar)|Pt(s)\)

Solution :

Step 1: Half-Cell Reactions and Overall Cell Reaction

Based on the cell notation $Sn(s)|Sn^{2+}(0.050 M)||H^{+}(0.020 M)|H_{2}(g) (1 bar)|Pt(s)$, the oxidation occurs at the anode and reduction at the cathode.

  • Anode (Oxidation): $Sn(s) \rightarrow Sn^{2+}(aq) + 2e^{-}$
  • Cathode (Reduction): $2H^{+}(aq) + 2e^{-} \rightarrow H_{2}(g)$

Adding the half-reactions to determine the overall cell reaction: [Per the law of conservation of charge]

$Sn(s) + 2H^{+}(aq) \rightarrow Sn^{2+}(aq) + H_{2}(g)$

The number of electrons involved in the transfer is $n = 2$.

Step 2: Determining Standard Cell Potential ($E^{\circ}_{cell}$)

Standard potentials at $298\ K$:

  • $E^{\circ}_{Sn^{2+}/Sn} = -0.14\ V$
  • $E^{\circ}_{H^{+}/H_{2}} = 0.00\ V$

$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$ [By definition of Standard EMF]

$E^{\circ}_{cell} = 0.00\ V - (-0.14\ V) = +0.14\ V$

Step 3: Formulating the Nernst Equation

The Nernst equation at $298\ K$ ($25^{\circ}C$) is given by: [Per the Nernst Equation derivation]

$E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Sn^{2+}][P_{H_{2}}]}{[H^{+}]^2}$

Step 4: Substitution and Numerical Calculation

Substitute the given concentrations and pressure: $[Sn^{2+}] = 0.050\ M$, $[H^{+}] = 0.020\ M$, $P_{H_{2}} = 1\ bar$.

$E_{cell} = 0.14 - \frac{0.0591}{2} \log \frac{0.050 \times 1}{(0.020)^2}$

$E_{cell} = 0.14 - 0.02955 \log \frac{0.050}{0.0004}$

$E_{cell} = 0.14 - 0.02955 \log(125)$

Using the logarithmic property $\log(125) \approx 2.0969$:

$E_{cell} = 0.14 - 0.02955 \times 2.0969$

$E_{cell} = 0.14 - 0.06196$

$E_{cell} = 0.07804\ V$

Visual Representation of Cell Components

Sn Electrode Pt/H2(g) Salt Bridge E = 0.078 V

Final Solution: The Nernst equation is $E_{cell} = 0.14 - \frac{0.0591}{2} \log \frac{[Sn^{2+}]}{[H^{+}]^2}$ and the EMF of the cell is $0.078\ V$.


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