Nernst Equation

Question

For the cell Zn|Zn²⁺(0.1M)||Cu²⁺(0.01M)|Cu at 298 K, if E°cell = 1.10 V, the cell potential is: (Take 2.303RT/F = 0.06)

RankGuru · Accepted Answer

Using the **Nernst Equation** for the cell reaction $Zn + Cu^{2+} 	o Zn^{2+} + Cu$ ($n=2$):
- Formula: $E_{cell} = E^\circ_{cell} - \frac{0.06}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$
- **Values**: $E^\circ_{cell} = 1.10 	ext{ V}$, $n = 2$, $[Zn^{2+}] = 0.1 	ext{ M}$, $[Cu^{2+}] = 0.01 	ext{ M}$.
- Calculation: $E_{cell} = 1.10 - \frac{0.06}{2} \log \left( \frac{0.1}{0.01} ight)$
$E_{cell} = 1.10 - 0.03 \log(10)$
$E_{cell} = 1.10 - 0.03(1) = 1.07 	ext{ V}$.

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