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

Solution :

Step 1: Identification of Half-Cell Reactions and Overall Cell Reaction

For the given cell notation: $Sn(s) | Sn^{2+}(0.050\,M) || H^+(0.020\,M) | H_2(g) (1\,bar) | Pt(s)$

  • Anode (Oxidation): $Sn(s) \rightarrow Sn^{2+}(aq) + 2e^-$ [Standard potential $E^\circ_{Sn^{2+}/Sn} = -0.14\,V$]
  • Cathode (Reduction): $2H^+(aq) + 2e^- \rightarrow H_2(g)$ [Standard potential $E^\circ_{H^+/H_2} = 0.00\,V$]

The overall cell reaction is:

$Sn(s) + 2H^+(aq) \rightarrow Sn^{2+}(aq) + H_2(g)$

The number of electrons transferred, $n = 2$.

Step 2: Calculation of Standard Cell Potential ($E^\circ_{cell}$)

$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ [Per the standard convention for electrode potentials]

$E^\circ_{cell} = 0.00\,V - (-0.14\,V) = 0.14\,V$

Step 3: Constructing the Nernst Equation

The Nernst equation for the reaction at $298\,K$ is given by:

$E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log \frac{[Sn^{2+}][P_{H_2}]}{[H^+]^2}$

Substituting the known values ($n=2, P_{H_2}=1\,bar$):

$E_{cell} = 0.14 - \frac{0.0591}{2} \log \frac{[0.050][1]}{[0.020]^2}$

Step 4: Algebraic Computation of Cell EMF

First, calculate the reaction quotient ($Q$):

$Q = \frac{0.050}{(0.020)^2} = \frac{0.050}{0.0004} = 125$

Next, determine the logarithmic value:

$\log(125) = \log(5^3) = 3 \times \log(5) \approx 3 \times 0.699 = 2.097$

Now, solve for $E_{cell}$:

$E_{cell} = 0.14 - 0.02955 \times 2.097$

$E_{cell} = 0.14 - 0.06196 \approx 0.078\,V$

Sn Electrode Pt/H2 Electrode Cell Diagram Representation

Step 5: Tabulated Summary

ParameterValue
$E^\circ_{cell}$$0.14\,V$
$n$$2$
$Q$$125$
$E_{cell}$$0.078\,V$

Final Solution: The Nernst equation is $E_{cell} = 0.14 - 0.02955 \log \frac{[Sn^{2+}]}{[H^+]^2}$ and the EMF of the cell is $0.078\,V$.


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