Question

Consider the following electrochemical cell at \(298\,\text{K}\): 
\[ \text{Pt} \, | \, \mathrm{HSnO_2^- (aq)} \, | \, \mathrm{Sn(OH)_6^{2-} (aq)} \, | \, \mathrm{OH^- (aq)} \, | \, \mathrm{Bi_2O_3 (s)} \, | \, \mathrm{Bi (s)} \] If the reaction quotient at a given time is \(10^6\), then the cell EMF (\(E_{\text{cell}}\)) is _________ \( \times 10^{-1} \) V (Nearest integer). 
Given: 
\[ E^\circ_{\mathrm{Bi_2O_3/Bi,OH^-}} = -0.44\ \text{V}, \quad E^\circ_{\mathrm{Sn(OH)_6^{2-}/HSnO_2^-,OH^-}} = -0.90\ \text{V} \]

Hint:

Always identify the number of electrons transferred correctly before applying the Nernst equation.

Answer 1

Correct Answer: 28 - 42

To determine the cell EMF (\(E_{\text{cell}}\)) for the given electrochemical cell at \(298\,\text{K}\), we use the Nernst Equation:

\[E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q\]

where:

• \(E^\circ_{\text{cell}}\) is the standard cell potential.
• \(R\) is the universal gas constant \((8.314 \, \text{J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1})\).
• \(T\) is the temperature in Kelvin \((298\,\text{K})\).
• \(n\) is the number of moles of electrons transferred in the balanced equation.
• \(F\) is Faraday’s constant \((96485 \, \text{C} \cdot \text{mol}^{-1})\).
• \(Q\) is the reaction quotient.

The given reaction quotient \(Q = 10^6\), and standard reduction potentials are:

• \(E^\circ_{\mathrm{Bi_2O_3/Bi,OH^-}} = -0.44 \, \text{V}\)
• \(E^\circ_{\mathrm{Sn(OH)_6^{2-}/HSnO_2^-,OH^-}} = -0.90 \, \text{V}\)

The standard cell potential \(E^\circ_{\text{cell}}\) is calculated as:

\[E^\circ_{\text{cell}} = E^\circ_{\mathrm{cathode}} - E^\circ_{\mathrm{anode}} = -0.44\, \text{V} - (-0.90\, \text{V}) = 0.46\, \text{V}\]

The balanced overall redox reaction involves 2 electrons, so \(n = 2\). Substitute these values into the Nernst Equation:

\[E_{\text{cell}} = 0.46 - \frac{8.314 \times 298}{2 \times 96485} \ln(10^6)\]

Simplify the equation:

\[E_{\text{cell}} = 0.46 - \frac{0.0257}{2} \times 13.815\quad(\text{using } \ln(10^6) \approx 13.815)\]

\[E_{\text{cell}} = 0.46 - 0.1775 \approx 0.2825\, \text{V}\]

Convert into nearest integer times \(10^{-1}\):

\[E_{\text{cell}} \approx 28 \times 10^{-1} \, \text{V}\]