Question

Calculate emf of the following cell at 298 K: \[ \text{Cr(s)} \, | \, \text{Cr}^{3+} (aq) \, (0.1\, M) \; || \; \text{Fe}^{2+} (aq) \, (0.01\, M) \, | \, \text{Fe(s)} \] Given: \[ E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74 \, V, \quad E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44 \, V, \quad \log 10 = 1 \]

Hint:

Steps for emf problems: 1. Identify anode and cathode 2. Balance electrons → find \( n \) 3. Calculate \( Q \) 4. Apply Nernst equation

Answer 1

Step 1: Write the Nernst Equation.
The Nernst equation is used to calculate the cell potential (emf) under non-standard conditions. It is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log Q \] where:
\( E \) = emf of the cell,
\( E^\circ \) = standard cell potential,
\( n \) = number of electrons transferred,
\( Q \) = reaction quotient (products/reactants).

Step 2: Determine the Cell Potential (E°).
The cell potential under standard conditions can be calculated using the standard reduction potentials of the half-reactions.
The given half-reactions and their standard electrode potentials are:
\[ \text{Cr(s)} + 3e^- \rightarrow \text{Cr}^{3+} \, (E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74 \, V) \] \[ \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe(s)} \, (E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44 \, V) \] The cell potential is calculated as: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] Since iron (Fe) is in the reduction half-reaction, it will be the cathode, and chromium (Cr) will be the anode.
\[ E^\circ_{\text{cell}} = (-0.44 \, V) - (-0.74 \, V) = +0.30 \, V \] Thus, the standard cell potential \( E^\circ_{\text{cell}} = +0.30 \, V \).

Step 3: Write the Reaction Quotient (Q).
The reaction is:
\[ \text{Cr(s)} + \text{Fe}^{2+} (aq) \rightarrow \text{Cr}^{3+} (aq) + \text{Fe(s)} \] The reaction quotient \( Q \) is:
\[ Q = \frac{[\text{Cr}^{3+}] [\text{Fe}]^2}{[\text{Fe}^{2+}] [\text{Cr}]^3} \] Substituting the concentrations:
\[ Q = \frac{(0.1)}{(0.01)} = 10 \] because the activities of pure solids are 1, and we are dealing with ions.

Step 4: Apply the Nernst Equation.
Now, using the Nernst equation, we can calculate the cell emf at 298 K:
\[ E = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q \] where \( n \) is the number of electrons transferred. Since the half-reactions involve 3 electrons for chromium and 2 electrons for iron, the total number of electrons transferred is the least common multiple of 2 and 3, which is 6.
\[ E = 0.30 - \frac{0.0591}{6} \log 10 \] Since \(\log 10 = 1\), the equation becomes: \[ E = 0.30 - \frac{0.0591}{6} \times 1 = 0.30 - 0.00985 = 0.29015 \, V \] Thus, the emf of the cell at 298 K is approximately 0.29 V.