Question

A galvanic cell consists of the following: \[ \text{Zn(s)} | \text{Zn}^{2+}(0.01M) || \text{Cu}^{2+}(0.1M) | \text{Cu(s)} \] The standard reduction potentials of the two electrodes are given as \( E^0(\text{Zn}^{2+}/\text{Zn}) = -0.763 \, \text{V} \) and \( E^0(\text{Cu}^{2+}/\text{Cu}) = 0.337 \, \text{V} \). The emf of the above cell will be:

• A. 1.13 V
• B. 1.50 V
• C. 0.455 V
• D. 1.10 V

Hint:

In a galvanic cell, the emf is the difference between the reduction potentials of the cathode and the anode.

Answer 1

The Correct Option is A

Step 1: Understanding the formula for emf.
The emf of the galvanic cell is given by: \[ E_{\text{cell}} = E^0(\text{cathode}) - E^0(\text{anode}) \] Here, the cathode is copper (\( \text{Cu}^{2+}/\text{Cu} \)) and the anode is zinc (\( \text{Zn}^{2+}/\text{Zn} \)). Step 2: Substituting the given values.
\[ E_{\text{cell}} = 0.337 - (-0.763) = 0.337 + 0.763 = 1.13 \, \text{V} \] Step 3: Conclusion.
The emf of the galvanic cell is \( \boxed{1.13} \, \text{V} \). The correct answer is (1) 1.13 V.

Answer 2

Since this cell is NOT at standard 1 M concentrations (\( \text{Zn}^{2+} \) is 0.01 M and \( \text{Cu}^{2+} \) is 0.1 M), the Nernst equation, not just the standard potentials, must be used. Let's check each option against a full Nernst calculation.

1. 1.13 V: The standard cell potential is \( E^0_{\text{cell}} = E^0(\text{Cu}^{2+}/\text{Cu}) - E^0(\text{Zn}^{2+}/\text{Zn}) = 0.337 - (-0.763) = 1.100 \) V. Applying the Nernst equation for \( n = 2 \): \( E = E^0 - \dfrac{0.0591}{2} \log \dfrac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = 1.100 - \dfrac{0.0591}{2} \log \dfrac{0.01}{0.1} = 1.100 - 0.02955 \times (-1) = 1.1296 \) V, which rounds to this value.
2. 1.50 V: Far higher than even the standard potential difference of 1.10 V; the concentration correction here is only a few hundredths of a volt, so it cannot push the emf up this much.
3. 0.455 V: Far too small - less than half the standard potential difference - and does not correspond to any reasonable concentration correction on top of 1.10 V.
4. 1.10 V: This is the STANDARD cell potential using 1 M concentrations for both half-cells; but the problem explicitly gives non-standard concentrations (0.01 M and 0.1 M), so the Nernst correction must be applied rather than stopping here.

Applying the Nernst equation correctly (not just the standard potentials) shifts the emf from 1.10 V up to about 1.13 V because the reaction quotient is less than 1.

Therefore, the correct answer is 1.13 V.