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Q2.5 (iv):
Write the Nernst equation and emf of the following cell at 298 K: (iv) (Pt(s)|Br^{-}(0.010 M)|Br_{2}(l )||H^{+}(0.030 M)| H_{2}(g) (1 bar)|Pt(s)).

Solution :

Step 1: Identification of Half-Cell Reactions

Based on the cell notation $Pt(s)|Br^{-}(0.010\,M)|Br_{2}(l)||H^{+}(0.030\,M)|H_{2}(g)(1\,bar)|Pt(s)$, the electrochemical cell consists of the following half-reactions:

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

Combining these, the overall cell reaction is:

$2Br^{-}(aq) + 2H^{+}(aq) \rightarrow Br_{2}(l) + H_{2}(g)$

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

Using standard electrode potentials from electrochemical series [Standard Hydrogen Electrode (SHE) $E^{\circ} = 0.00\,V$; $Br_{2}/Br^{-} = 1.09\,V$]:

$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$

$E^{\circ}_{cell} = E^{\circ}_{(H^{+}/H_{2})} - E^{\circ}_{(Br_{2}/Br^{-})}$

$E^{\circ}_{cell} = 0.00\,V - 1.09\,V = -1.09\,V$

Step 3: Nernst Equation Formulation

The Nernst equation at $298\,K$ for a cell involving $n$ electrons is given by [Per the Nernst Equation]:

$E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log Q$

For the given reaction, $n = 2$. The reaction quotient $Q$ is expressed as:

$Q = \frac{[H_{2}][Br_{2}]}{[H^{+}]^{2}[Br^{-}]^{2}}$

Since $Br_{2}$ is a liquid, its activity is $1$. Given $P_{H_{2}} = 1\,bar$, $[H^{+}] = 0.030\,M$, and $[Br^{-}] = 0.010\,M$:

$Q = \frac{1}{(0.030)^{2} \times (0.010)^{2}} = \frac{1}{9 \times 10^{-4} \times 1 \times 10^{-4}} = \frac{1}{9 \times 10^{-8}} = 1.11 \times 10^{7}$

Step 4: Calculation of Emf of the Cell

$E_{cell} = -1.09 - \frac{0.0591}{2} \log(1.11 \times 10^{7})$

$E_{cell} = -1.09 - 0.02955 \times (\log 1.11 + 7)$

$E_{cell} = -1.09 - 0.02955 \times (0.0453 + 7)$

$E_{cell} = -1.09 - 0.02955 \times (7.0453)$

$E_{cell} = -1.09 - 0.2082 \approx -1.298\,V$

Br⁻ (0.01M) H⁺ (0.03M) Anode (-) Cathode (+)

Final Solution: The Nernst equation is $E_{cell} = -1.09 - \frac{0.0591}{2} \log \frac{1}{[H^{+}]^{2}[Br^{-}]^{2}}$ and the emf of the cell is $-1.298\,V$.


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