To find the pH of the solution using the quinhydrone electrode and the saturated calomel electrode, we can use the Nernst equation which relates the cell potential (E) to the standard electrode potentials and the concentration of hydrogen ions in the solution.
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Identifying the Electrode Reactions:
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For the quinhydrone electrode: \[ \text{Quinhydrone (QH)} + \text{H}^+ \rightleftharpoons \text{QH}2 \] The standard electrode potential for this reaction is given as \( E^\circ{QH} = 0.6996 , \text{V} \).
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For the saturated calomel electrode (SCE, which measures the potential of the hydrogen electrode): \[ \text{Hg}_2\text{Cl}2 + 2\text{e}^- \rightleftharpoons 2\text{Hg} + 2\text{Cl}^- \] The standard electrode potential is given as \( E^\circ{SCE} = 0.2422 , \text{V} \).
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Calculating the Cell Potential: The overall cell potential \( E \) can be expressed as: \[ E = E_{QH} - E_{SCE} \] Plugging in the known values: \[ E = 0.1520 , \text{V} \] \( E_{QH} \) can be expressed in terms of its standard potential and the concentration of hydrogen ions: \[ E_{QH} = E^\circ_{QH} - \frac{0.0591}{n} \log[\text{H}^+] \] Where:
- \( n = 1 \) (one electron is involved in the quinhydrone electrode reaction).
So, \[ E_{QH} = 0.6996 - 0.0591 \log[\text{H}^+] \]
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Combining the Equations: The cell potential can also be defined as: \[ E = E_{QH} - E_{SCE} \] Substituting for \( E_{QH} \): \[ 0.1520 = (0.6996 - 0.0591 \log[\text{H}^+]) - 0.2422 \] Simplifying: \[ 0.1520 = 0.4574 - 0.0591 \log[\text{H}^+] \] Re-arranging gives: \[ 0.0591 \log[\text{H}^+] = 0.4574 - 0.1520 \] \[ 0.0591 \log[\text{H}^+] = 0.3054 \] \[ \log[\text{H}^+] = \frac{0.3054}{0.0591} \] \[ \log[\text{H}^+] \approx 5.166 \]
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Calculating \([\text{H}^+]\): The concentration of hydrogen ions: \[ [\text{H}^+] \approx 10^{-5.166} \approx 6.76 \times 10^{-6} , \text{M} \]
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Calculating the pH: \[ \text{pH} = -\log[\text{H}^+] \approx -\log(6.76 \times 10^{-6}) \approx 5.17 \]
So, the pH of the solution is approximately 5.17.
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