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Q2.5(ii):
Write the Nernst equation and emf of the following cell at 298 K: \(Fe(s)|Fe^{2+}(0.001M)||H^{+}(1M)|H_{2}(g)(1bar)| Pt(s)\)

Solution :

Step 1: Cell Reaction and Potential Determination

The given electrochemical cell is: $Fe(s) | Fe^{2+}(0.001 M) || H^{+}(1 M) | H_{2}(g)(1 bar) | Pt(s)$

The half-cell reactions are defined as follows:

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

The number of electrons transferred, $n = 2$ [Per the stoichiometry of the balanced redox reaction].

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

Using standard reduction potentials ($E^{\circ}$):

  • $E^{\circ}_{H^{+}/H_{2}} = 0.00 \, V$
  • $E^{\circ}_{Fe^{2+}/Fe} = -0.44 \, V$

$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$ [By the standard IUPAC convention for calculating EMF]

$E^{\circ}_{cell} = 0.00 \, V - (-0.44 \, V) = 0.44 \, V$

Step 3: The Nernst Equation Setup

The Nernst equation at $298 \, K$ is expressed as:

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

Where $Q$ (Reaction Quotient) is defined as: $Q = \frac{[Fe^{2+}] \cdot P_{H_{2}}}{[H^{+}]^{2}}$ [Based on the Law of Mass Action for the cell reaction].

Step 4: Calculation of EMF

Substituting the given concentrations: $[Fe^{2+}] = 10^{-3} \, M$, $[H^{+}] = 1 \, M$, and $P_{H_{2}} = 1 \, bar$.

$Q = \frac{10^{-3} \times 1}{(1)^{2}} = 10^{-3}$

$E_{cell} = 0.44 - \frac{0.0591}{2} \log(10^{-3})$

$E_{cell} = 0.44 - 0.02955 \times (-3)$

$E_{cell} = 0.44 + 0.08865$

$E_{cell} = 0.52865 \, V$

Fe(s) Pt(s) Fe2+ (0.001M) || H+ (1M), H2 (1bar)

Final Solution: The EMF of the cell is 0.52865 V.


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