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: Standard cell potential: \( E^0_{\text{cell}} = 0.337 - (-0.763) = 1.100 \) V.
Step 2: Since concentrations are non-standard, apply the Nernst equation with \( n = 2 \): \( E = E^0 - \dfrac{0.0591}{2} \log \dfrac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \).
Step 3: \( \log \dfrac{0.01}{0.1} = \log(0.1) = -1 \), so \( E = 1.100 - \dfrac{0.0591}{2}(-1) = 1.100 + 0.0296 \).
\[ \boxed{E \approx 1.13 \ \text{V}} \]

Answer 2

Rather than computing the full Nernst correction first, let's reason about its direction and rough size, then match that against the options.

1. 1.13 V: The reaction quotient \( Q = \dfrac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \dfrac{0.01}{0.1} = 0.1 \) is less than 1, so \( \log Q \) is negative, and since the Nernst term is subtracted with a negative sign, the correction adds a small positive amount to the standard 1.10 V, nudging it up to about 1.13 V - consistent with a modest, sub-0.1-volt increase.
2. 1.50 V: Would need a Nernst correction of about 0.4 V, far larger than the roughly 0.03 V correction this reaction quotient can produce.
3. 0.455 V: Would require the correction to swing the potential down by more than half a volt, the wrong direction and far too large given \( Q \) is close to 1 in magnitude.
4. 1.10 V: This ignores the Nernst correction entirely and reports only the standard potential, which does not account for the given non-standard concentrations.

A small, positive correction on top of the 1.10 V standard potential (because \( Q \textless 1 \)) points to a final emf just above 1.10 V.

Therefore, the correct answer is 1.13 V.