Question

Consider the following cell: $ \text{Pt}(s) \, \text{H}_2 (1 \, \text{atm}) | \text{H}^+ (1 \, \text{M}) | \text{Cr}_2\text{O}_7^{2-}, \, \text{Cr}^{3+} | \text{H}^+ (1 \, \text{M}) | \text{Pt}(s) $ 
Given: $ E^\circ_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} = 1.33 \, \text{V}, \quad \left[ \text{Cr}^{3+} \right]^2 / \left[ \text{Cr}_2\text{O}_7^{2-} \right] = 10^{-7} $ 
At equilibrium: $ \left[ \text{Cr}^{3+} \right]^2 / \left[ \text{Cr}_2\text{O}_7^{2-} \right] = 10^{-7} $ 
Objective: $ \text{Determine the pH at the cathode where } E_{\text{cell}} = 0. $

Hint:

For electrochemical cells, the Nernst equation is crucial for relating cell potential to concentration changes. Remember that at equilibrium, the cell potential becomes zero, and you can use this to solve for unknown values like pH or concentration.

Answer 1

Correct Answer: 10 - 11

Given the cell: \[ \text{Pt}|\text{H}_2(1\,\text{atm})|\text{H}^+(1\,\text{M})||\text{Cr}_2\text{O}_7^{2-},\text{Cr}^{3+},\text{H}^+|\text{Pt} \] with \( E^\circ_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} = 1.33\,\text{V} \) and equilibrium condition: \[ \frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}]} = 10^{-7} \]
Steps: 1.
Half-reactions: \begin{itemize} \item Anode: \(\text{H}_2 \rightarrow 2\text{H}^+ + 2e^-\) \quad (\(E^\circ = 0.00\,\text{V}\)) \item Cathode: \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\) \end{itemize} 2.
Nernst equations: \begin{itemize} \item Anode: \(E_{\text{anode}} = 0.00 - \frac{0.0591}{2}\log(1) = 0.00\,\text{V}\) \item Cathode: \[ E_{\text{cathode}} = 1.33 - \frac{0.0591}{6}\log\left(\frac{10^{-7}}{[\text{H}^+]^{14}}\right) \] \[ = 1.33 + 0.069 + 0.138\log[\text{H}^+] \] \[ = 1.399 - 0.138\,\text{pH} \] \end{itemize} 3.
Cell potential (\(E_{\text{cell}} = 0\)): \[ 0 = (1.399 - 0.138\,\text{pH}) - 0 \] \[ \text{pH} = \frac{1.399}{0.138} \approx 10.14 \] \subsection{Answer:} The required pH is \(\boxed{10.14}\).