Question

At 298 K the emf of the cell given below is 0.87 V.
Pt, \( \text{H}_2(1 \text{ atm})|\text{H}^+(\text{aq}) || \text{Ag}^+(1 \text{ M})|\text{Ag}(\text{s}) \)
The pH of the acid solution is
(Given : \( \text{E}^{\circ}_{\text{Ag}^+|\text{Ag}} = 0.80 \text{ V}; \text{E}^{\circ}_{2\text{H}^+|\text{H}_2} = 0.0 \text{ V}) \)

• A. 2.18
• B. 3.18
• C. 1.18
• D. 1.09

Hint:

When using the Nernst equation, correctly identify the anode (oxidation) and cathode (reduction) to calculate $\text{E}^\circ_{\text{cell}}$. Pay careful attention to the number of electrons transferred ($n$) and the form of the reaction quotient ($Q$). Remember that $\text{pH} = -\log[\text{H}^+]$. For problems involving concentration of ions, make sure to include them in the reaction quotient correctly raised to their stoichiometric coefficients.

Answer 1

The Correct Option is C

Step 1: Identify Anode and Cathode Half-Reactions and Write the Overall Cell Reaction
The cell notation is Pt, \( \text{H}_2(1 \text{ atm})|\text{H}^+(\text{aq}) || \text{Ag}^+(1 \text{ M})|\text{Ag}(\text{s}) \). The double vertical lines ($||$) represent the salt bridge. The left side is the anode (oxidation) and the right side is the cathode (reduction). \begin{itemize} \item Anode (Oxidation): Hydrogen electrode \[ \text{H}_2(\text{g}) \rightarrow 2\text{H}^+(\text{aq}) + 2e^- \] The standard reduction potential for this half-reaction is given as $\text{E}^{\circ}_{2\text{H}^+|\text{H}_2} = 0.0 \text{ V}$. \item Cathode (Reduction): Silver electrode \[ \text{Ag}^+(\text{aq}) + e^- \rightarrow \text{Ag}(\text{s}) \] The standard reduction potential for this half-reaction is given as $\text{E}^{\circ}_{\text{Ag}^+|\text{Ag}} = 0.80 \text{ V}$. \end{itemize} To balance the electrons, multiply the cathode reaction by 2: \[ 2\text{Ag}^+(\text{aq}) + 2e^- \rightarrow 2\text{Ag}(\text{s}) \] Overall Cell Reaction: Add the balanced half-reactions: \[ \text{H}_2(\text{g}) + 2\text{Ag}^+(\text{aq}) \rightarrow 2\text{H}^+(\text{aq}) + 2\text{Ag}(\text{s}) \] Step 2: Calculate the Standard Cell Potential (\( \text{E}^{\circ}_{\text{cell}} \))
\[ \text{E}^{\circ}_{\text{cell}} = \text{E}^{\circ}_{\text{cathode}} - \text{E}^{\circ}_{\text{anode}} \] \[ \text{E}^{\circ}_{\text{cell}} = \text{E}^{\circ}_{\text{Ag}^+|\text{Ag}} - \text{E}^{\circ}_{\text{H}^+|\text{H}_2} \] \[ \text{E}^{\circ}_{\text{cell}} = 0.80 \text{ V} - 0.0 \text{ V} = 0.80 \text{ V} \] Step 3: Apply the Nernst Equation
The Nernst equation at 298 K is: \[ \text{E}_{\text{cell}} = \text{E}^{\circ}_{\text{cell}} - \frac{0.0592}{n} \log Q \] Where: \begin{itemize} \item $\text{E}_{\text{cell}}$ is the measured cell potential ($0.87 \text{ V}$). \item $\text{E}^{\circ}_{\text{cell}}$ is the standard cell potential ($0.80 \text{ V}$). \item $n$ is the number of moles of electrons transferred in the balanced reaction ($n=2$). \item $Q$ is the reaction quotient. \end{itemize} For the reaction $\text{H}_2(\text{g}) + 2\text{Ag}^+(\text{aq}) \rightarrow 2\text{H}^+(\text{aq}) + 2\text{Ag}(\text{s})$, the reaction quotient $Q$ is: \[ Q = \frac{[\text{H}^+]^2 \times (\text{activity of Ag(s)})^2}{(\text{partial pressure of H}_2) \times [\text{Ag}^+]^2} \] Since the activity of pure solids is 1 and the partial pressure of $\text{H}_2$ is $1 \text{ atm}$: \[ Q = \frac{[\text{H}^+]^2}{1 \times [\text{Ag}^+]^2} = \frac{[\text{H}^+]^2}{[\text{Ag}^+]^2} \] Given $[\text{Ag}^+] = 1 \text{ M}$: \[ Q = \frac{[\text{H}^+]^2}{(1)^2} = [\text{H}^+]^2 \] Substitute the values into the Nernst equation: \[ 0.87 = 0.80 - \frac{0.0592}{2} \log ([\text{H}^+]^2) \] \[ 0.87 - 0.80 = - \frac{0.0592}{2} (2 \log [\text{H}^+]) \] \[ 0.07 = - 0.0592 \log [\text{H}^+] \] Step 4: Relate \( \log [\text{H}^+] \) to pH and Calculate pH
Recall that $\text{pH} = -\log [\text{H}^+]$. So, the equation becomes: \[ 0.07 = 0.0592 \times \text{pH} \] Solve for pH: \[ \text{pH} = \frac{0.07}{0.0592} \] \[ \text{pH} \approx 1.1824 \] Rounding to two decimal places, $\text{pH} \approx 1.18$. Step 5: Analyze Options
\begin{itemize} \item Option (1): 2.18. Incorrect. \item Option (2): 3.18. Incorrect. \item Option (3): 1.18. Correct, as it matches our calculated pH. \item Option (4): 1.09. Incorrect. \end{itemize}