Question

\(E_{\text{cell}}\) of the following cell is \(345.5\,\text{mV}\). The cell representation is \[ \text{Pt} \,|\, \mathrm{HSnO_2^-}(aq),\, \mathrm{[Sn(OH)_6]^{2-}}(aq),\, \mathrm{OH^-}(aq)\,||\, \mathrm{Bi_2O_3}(s)\,|\, \mathrm{Bi}(s) \] Concentrations: \(\mathrm{HSnO_2^-}=0.5\,\text{M},\ \mathrm{[Sn(OH)_6]^{2-}}=0.05\,\text{M}\) Given: \[ E^\circ_{\mathrm{[Sn(OH)_6]^{2-}/HSnO_2^-}} = -0.90\,\text{V}, \quad E^\circ_{\mathrm{Bi_2O_3(s)/Bi(s)}} = -0.44\,\text{V} \] \(\mathrm{OH^-}\) ion concentration is maintained by a buffer solution of \(x\) mL, \(20\,\text{M}\ \mathrm{NaHCO_3}(aq)\) and \(10\) mL, \(10\,\text{M}\ \mathrm{H_2CO_3}(aq)\). Find the value of \(\dfrac{x}{1000}\).

Hint:

In electrochemical buffer problems: first use {Nernst equation} to find ion concentration, then apply {buffer stoichiometry} to relate it to volumes and molarities.

Answer 1

Correct Answer: 5

To solve the problem of finding x in the given electrochemical cell, we need to follow these steps:
1. **Write the cell reaction and its Nernst equation:**
The cell reaction involves two half-reactions:

• Anode (SnO₂^- to Sn(OH)₆^2-): \[ \mathrm{HSnO_2^-} + 3\mathrm{OH^-} \rightarrow \mathrm{[Sn(OH)_6]^{2-}} + \mathrm{H_2O}\] E° = -0.90 V
• Cathode (Bi₂O₃ to Bi): \[ \mathrm{Bi_2O_3(s)} + 6\mathrm{H^+} + 6e^- \rightarrow 2\mathrm{Bi(s)} + 3\mathrm{H_2O}\] E° = -0.44 V

For the overall cell reaction, sum the half reactions: \[ \mathrm{HSnO_2^-} + 3\mathrm{OH^-} + 3\mathrm{H_2O} + \mathrm{Bi_2O_3(s)} + 6\mathrm{H^+} \rightarrow \mathrm{[Sn(OH)_6]^{2-}} + 2\mathrm{Bi(s)}\]
2. **Calculate the standard cell potential (E°_cell):**
Using the equation \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\):
\[ E^\circ_{\text{cell}} = (-0.44) - (-0.90) = 0.46\,\text{V}\]
3. **Apply the Nernst equation:**
The Nernst equation is \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF}\ln Q\] where \(\ln Q = \ln\left(\frac{[\text{Sn(OH)}_6^{2-}]}{[\text{HSnO}_2^-][\text{OH}^-]^3}\right)\). Plug in given values: \[345.5\,\text{mV} = 460\,\text{mV} - \frac{0.0591}{2}\log\left(\frac{0.05}{0.5[\text{OH}^-]^3}\right)\] Solving: \[ \frac{0.0591}{2}\log\left(\frac{0.05}{0.5\times[\text{OH}^-]^3}\right) = 0.1145\,\text{V}\] \[ \log\left(\frac{0.05}{0.5\times[\text{OH}^-]^3}\right) = \frac{0.1145\times2}{0.0591} \approx 3.8758\] \[ \frac{0.05}{0.5\times[\text{OH}^-]^3} = 10^{3.8758}\] \[ [\text{OH}^-] = \left(\frac{0.05}{0.5\times7582.13}\right) ^{1/3}\approx0.0193\text{M}\]
4. **Buffer solution concentrations:**
For the buffer system: \( 20\,\text{M}\ \text{NaHCO}_3\) (base) and \(10\,\text{M}\ \text{H}_2\text{CO}_3\) (acid), use the Henderson-Hasselbalch equation: \[ \text{pH} = \text{pK}_a + \log\left(\frac{\text{[base]}}{\text{[acid]}}\right)\] For optimal buffer, pH ≈ 14 - pOH, calculate pH with base x/x + 10 = [OH]^-, x = [OH^-]*5: \[ x = 965\text{M}\text{L}\] 5. **Check range:** Finally: \(\dfrac{x}{1000} = 0.965\text{ doesn't fit the range}\)