The measured EMF of the concentration cell in which the following reaction occurs: is found to be at — Electrochemistry Chemistry Question
Question
The measured EMF of the concentration cell in which the following reaction occurs: $Zn(s) + Ni^{2+}(aq, 1.0\text{ M}) \rightleftharpoons Zn^{2+}(aq, 10.0\text{ M}) + Ni(s)$ is found to be $0.5105\text{ V}$ at $298\text{ K}$. Calculate the standard EMF ($E^\circ_{cell}$) of the cell in Volts. (Use $\frac{2.303RT}{F} = 0.059\text{ V}$)
Answer: 0.54
💡 Solution & Explanation
Using the Nernst equation: $E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Ni^{2+}]}$. Here $n = 2$. $0.5105 = E^\circ_{cell} - \frac{0.059}{2} \log(10/1.0)$. $0.5105 = E^\circ_{cell} - 0.0295 \times \log(10)$. $0.5105 = E^\circ_{cell} - 0.0295$. $E^\circ_{cell} = 0.5105 + 0.0295 = 0.54\text{ V}$.
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