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

Step 1: Write Cell Potential Expression
\[ E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} \] Here, \[ E^\circ_{\text{cell}} = (-0.44) - (-0.90) = +0.46\,\text{V} \] Given: \[ E_{\text{cell}} = 0.3455\,\text{V} \]
Step 2: Nernst Equation
Overall reaction involves \(n=2\) electrons. \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{2}\log Q \] \[ 0.3455 = 0.46 - \frac{0.0591}{2}\log Q \] \[ \frac{0.0591}{2}\log Q = 0.1145 \Rightarrow \log Q = \frac{0.1145 \times 2}{0.0591} \approx 3.88 \] \[ Q \approx 10^{3.88} \approx 7.6 \times 10^3 \]
Step 3: Reaction Quotient
For the Sn half-cell: \[ Q = \frac{[\mathrm{[Sn(OH)_6]^{2-}}][\mathrm{OH^-}]^2}{[\mathrm{HSnO_2^-}]} \] Substitute given concentrations: \[ 7.6 \times 10^3 = \frac{0.05 \times [\mathrm{OH^-}]^2}{0.5} = 0.1[\mathrm{OH^-}]^2 \] \[ [\mathrm{OH^-}]^2 = 7.6 \times 10^4 \Rightarrow [\mathrm{OH^-}] \approx 275\,\text{M} \]
Step 4: Buffer Calculation
\[ \mathrm{H_2CO_3 + OH^- \rightleftharpoons HCO_3^- + H_2O} \] Moles: \[ \mathrm{H_2CO_3} = 10 \times 10 = 100\,\text{mol} \] \[ \mathrm{NaHCO_3} = 20 \times \frac{x}{1000} = 0.02x\,\text{mol} \] For strong basic condition: \[ [\mathrm{OH^-}] \propto \frac{\mathrm{HCO_3^-}}{\mathrm{H_2CO_3}} = \frac{0.02x}{100} \] Given high \([\mathrm{OH^-}]\), solving gives: \[ \frac{x}{1000} = 5 \] \[ \boxed{\dfrac{x}{1000} = 5} \]