Question

For $Zn(s)+Cu^{2+}(0.1M) \rightarrow Zn^{2+}(0.001 M)+Cu(s)$, calculate $E_{cell}$ if $E^0_{cell} = 1.1V$. ________.

• A. 1.218 V
• B. 1.118 V
• C. 1.159 V
• D. 1.041 V
• E. 0.982 V

Hint:

If product concentration $<$ reactant concentration, $E_{cell} > E^0_{cell}$.

Answer 1

The Correct Option is C

Step 1: Concept
Use the Nernst Equation: $E_{cell} = E^0_{cell} - \frac{0.059}{n} \log Q$.

Step 2: Meaning
$n = 2$ ($Zn \rightarrow Zn^{2+} + 2e^-$). $Q = \frac{[Zn^{2+}]}{[Cu^{2+}]} = \frac{0.001}{0.1} = 10^{-2}$.

Step 3: Analysis
$E_{cell} = 1.1 - \frac{0.059}{2} \log(10^{-2}) = 1.1 - (0.0295 \times -2)$. $E_{cell} = 1.1 + 0.059$.

Step 4: Conclusion
$E_{cell} = 1.159 V$. Final Answer: (C)