A standard galvanic cell is constructed as follows: . The observed EMF of the cell is at . Given tha — Electrochemistry Chemistry Question
Question
A standard galvanic cell is constructed as follows: $Pt, H_2(1\text{ atm}) \| H^+(pH = x) \|\| \text{Normal Calomel Electrode}$. The observed EMF of the cell is $0.67\text{ V}$ at $25^\circ\text{C}$. Given that the standard reduction potential of the normal calomel electrode is $0.28\text{ V}$, what is the value of $x$? (Use $\frac{2.303RT}{F} = 0.059\text{ V}$)
Answer: A
💡 Solution & Explanation
The cell potential is $E_{cell} = E_{cathode} - E_{anode}$. The cathode is the calomel electrode ($0.28\text{ V}$). The anode is the hydrogen electrode. $E_{anode} = E^\circ_{H^+/H_2} - \frac{0.059}{1} \log \frac{1}{[H^+]} = 0 - 0.059(pH) = -0.059x$. Therefore, $0.67 = 0.28 - (-0.059x) = 0.28 + 0.059x$. Solving for $x$ gives $x = \frac{0.39}{0.059} \approx 6.61$.
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