Question

\(E_{cell}\), for the cell given below is 0.82 V. What is its \(E^{\circ}\) value? \(Fe | Fe^{2+}(0.001 \text{ M}) || Cu^{2+}(0.1 \text{ M}) | Cu\). (Given: \(\frac{2.303 RT}{F} = 0.06 \text{ V}\))

• A. 0.63 V
• B. 0.69 V
• C. 0.76 V
• D. 0.87 V

Hint:

Always ensure the value of 'n' matches the number of electrons involved in the balanced redox equation.

Answer 1

The Correct Option is C

Concept: The Nernst equation relates the cell potential to the standard cell potential and the reaction quotient: \[ E_{cell} = E^{\circ}_{cell} - \frac{0.059}{n} \log Q \]

Step 1: Write the cell reaction and determine \(n\).
The cell reaction is: \[ Fe(s) + Cu^{2+}(aq) \rightarrow Fe^{2+}(aq) + Cu(s) \] Here, \(n = 2\) electrons are transferred.

Step 2: Calculate the reaction quotient \(Q\).
\[ Q = \frac{[Fe^{2+}]}{[Cu^{2+}]} = \frac{0.001}{0.1} = 0.01 = 10^{-2} \]

Step 3: Apply the Nernst equation.
Given \(\frac{0.06}{n} \approx \frac{0.06}{2} = 0.03\): \[ 0.82 = E^{\circ}_{cell} - 0.03 \log(10^{-2}) \] \[ 0.82 = E^{\circ}_{cell} - 0.03 \times (-2) \] \[ 0.82 = E^{\circ}_{cell} + 0.06 \] \[ E^{\circ}_{cell} = 0.82 - 0.06 = 0.76 \text{ V} \] \[ \boxed{0.76 \text{ V}} \]