A cell is constructed as . The observed EMF is at . Given that , what is the approximate pH of the w — Electrochemistry Chemistry Question
Question
A cell is constructed as $Pt \| H_2(1\text{ atm}) \| HOCN(1.3 \times 10^{-3}\text{ M}) \|\| Ag^+(0.8\text{ M}) \| Ag(s)$. The observed EMF is $0.982\text{ V}$ at $298\text{ K}$. Given that $E^\circ_{Ag^+/Ag} = 0.80\text{ V}$, what is the approximate pH of the weak $HOCN$ solution located in the anode compartment? (Use $\frac{2.303RT}{F} = 0.059\text{ V}$)
💡 Solution & Explanation
The overall cell reaction is $\frac{1}{2}H_2 + Ag^+ \rightarrow H^+ + Ag$. The standard cell potential is $E^\circ_{cell} = 0.80 - 0.00 = 0.80\text{ V}$. Applying the Nernst equation: $E_{cell} = E^\circ_{cell} - 0.059 \log \frac{[H^+]}{[Ag^+]}$. Substituting the knowns: $0.982 = 0.80 - 0.059 \log \frac{[H^+]}{0.8}$, simplifying to $0.182 = -0.059 (\log[H^+] - \log 0.8)$. Therefore, $\frac{0.182}{-0.059} \approx -3.085 = \log[H^+] - \log(0.8)$. Since $\log(0.8) \approx -0.097$, we get $-3.085 = \log[H^+] + 0.097$, so $\log[H^+] \approx -3.182$. Taking the negative gives $pH \approx 3.18$, which approximates closely to 3.2.