Question

For the cell reaction \[ Zn(s)+Cu^{2+}(aq)\rightarrow Zn^{2+}(aq)+Cu(s) \] the standard cell potential is \(1.10\ V\). If \[ [Zn^{2+}] = 0.10\ M \] and \[ [Cu^{2+}] = 1.0\times10^{-3}\ M \] at \(298\ K\), the cell potential is closest to:

• A. \(0.98\ V\)
• B. \(1.04\ V\)
• C. \(1.16\ V\)
• D. \(1.22\ V\)

Hint:

If \[ Q>1 \] then \[ E_{cell}<E^\circ_{cell} \] If \[ Q<1 \] then \[ E_{cell}>E^\circ_{cell} \] This shortcut helps eliminate options quickly.

Answer 1

The Correct Option is B

Concept: This is a direct application of the Nernst Equation. For a reaction involving \(n\) electrons: \[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \] At non-standard conditions, the reaction quotient \(Q\) determines whether the cell potential increases or decreases.

Step 1: Determine the number of electrons transferred. Reaction: \[ Zn \rightarrow Zn^{2+}+2e^- \] \[ Cu^{2+}+2e^- \rightarrow Cu \] Thus: \[ n=2 \]

Step 2: Calculate the reaction quotient. \[ Q = \frac{[Zn^{2+}]}{[Cu^{2+}]} \] Substituting: \[ = \frac{0.10}{10^{-3}} \] \[ = 100 \] \[ =10^2 \]

Step 3: Apply Nernst Equation. \[ E = 1.10 - \frac{0.0591}{2} \log(100) \] Since: \[ \log100=2 \] Therefore: \[ E = 1.10 - \frac{0.0591}{2}\times2 \] \[ = 1.10-0.0591 \] \[ = 1.0409 \] \[ \approx1.04V \]

Step 4: Interpretation. Since: \[ Q>1 \] the cell potential decreases from its standard value. Hence: \[ E<E^\circ \] which agrees with our result.

Step 5: Final conclusion. \[ \boxed{E_{cell}=1.04V} \] Therefore: \[ \boxed{\text{Option (B)}} \]