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

Solution :

Step 1: Identification of Half-Cell Reactions

Based on the cell notation $Pt(s)|Br^{-}(0.010 M)|Br_{2}(l )||H^{+}(0.030 M)| H_{2}(g) (1 bar)|Pt(s)$, we define the oxidation and reduction half-reactions:

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

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

[As per the IUPAC convention for electrochemical cells, oxidation occurs at the anode (left) and reduction occurs at the cathode (right).]

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

The standard potential is determined by the difference between the standard reduction potentials of the cathode and the anode:

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

Given standard electrode potentials from electrochemical series:

  • $E^{\circ}_{H^{+}/H_{2}} = 0.00 \, V$
  • $E^{\circ}_{Br_{2}/Br^{-}} = +1.09 \, V$
$E^{\circ}_{cell} = 0.00 \, V - 1.09 \, V = -1.09 \, V$

Step 3: The Nernst Equation Formulation

For the reaction $2Br^{-}(aq) + 2H^{+}(aq) \rightarrow Br_{2}(l) + H_{2}(g)$, the number of electrons transferred ($n$) is 2. The Nernst equation at $298 \, K$ is:

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

The reaction quotient $Q$ is expressed as:

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

[Since $Br_{2}(l)$ is a pure liquid, its activity is 1. The pressure of $H_{2}(g)$ is given as 1 bar.]

$Q = \frac{1 \times 1}{(0.010)^{2} \times (0.030)^{2}}$

Step 4: Numerical Computation

Calculate $Q$:

$Q = \frac{1}{(1.0 \times 10^{-2})^{2} \times (3.0 \times 10^{-2})^{2}}$ $Q = \frac{1}{(1.0 \times 10^{-4}) \times (9.0 \times 10^{-4})} = \frac{1}{9.0 \times 10^{-8}} = 1.11 \times 10^{7}$

Substitute values into the Nernst Equation:

$E_{cell} = -1.09 - \frac{0.0591}{2} \log(1.11 \times 10^{7})$ $E_{cell} = -1.09 - 0.02955 \times (\log(1.11) + 7)$ $E_{cell} = -1.09 - 0.02955 \times (0.0453 + 7)$ $E_{cell} = -1.09 - 0.02955 \times (7.0453)$ $E_{cell} = -1.09 - 0.2082$ $E_{cell} = -1.2982 \, V$

Visual Representation of Cell Potential

Standard Potential (-1.09V) Actual Potential (-1.298V)

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


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