Question

If E$_{cell}$ of the following reaction is x $\times$ 10$^{-1}$. Find x 
\(\text{Pt/ HSnO$_2$ / Sn(OH)$_6^{2-}$, OH$^-$ / Bi$_2$O$_3$ / Bi / Pt}\)
\(\text{[Reaction Quotient, Q = 10$^6$]}\)
Given \( E^o_{\text{[Sn(OH)$_3$]}} \) = -0.90 V, \( E^o_{\text{Bi$_2$O$_3$ / Bi}} \) = -0.44 V 
 

Hint:

When using the Nernst equation, remember that the reaction quotient \(Q\) and the number of electrons involved (\(n\)) are key to calculating the cell potential at non-standard conditions.

Answer 1

Correct Answer: 4

Step 1: Calculate $\Delta G^\circ_{reaction$.} Reaction: \(A_2 \rightleftharpoons 2A\) \[ \Delta G^\circ_{rxn} = 2\Delta G^\circ(A) - \Delta G^\circ(A_2) \] \[ \Delta G^\circ_{rxn} = 2(-50.8625) - (-100) = -101.725 + 100 = -1.725\,\text{kJ/mol} = -1725\,\text{J/mol} \]
Step 2: Calculate Equilibrium Constant K. \[ \Delta G^\circ = -RT \ln K \] \[ -1725 = -(8.3)(300) \ln K \] \[ \ln K = \frac{1725}{2490} \approx 0.693 \] \[ K \approx e^{0.693} \approx 2 \]
Step 3: Relate K to degree of dissociation $\alpha$. For \(A_2 \rightleftharpoons 2A\) starting with 1 mole at 1 atm pressure: Total moles at equilibrium = \(1 - \alpha + 2\alpha = 1 + \alpha\). Partial pressures: \(p_{A2} = \frac{1-\alpha}{1+\alpha}P\), \(p_A = \frac{2\alpha}{1+\alpha}P\). \[ K_p = \frac{(p_A)^2}{p_{A2}} = \frac{4\alpha^2 P}{(1-\alpha)(1+\alpha)} = \frac{4\alpha^2}{1-\alpha^2} \] Since K = 2: \[ 2 = \frac{4\alpha^2}{1-\alpha^2} \implies 1-\alpha^2 = 2\alpha^2 \implies 3\alpha^2 = 1 \] \[ \alpha = \frac{1}{\sqrt{3}} \approx 0.577 \]
Step 4: Format the answer. \(\alpha = 57.7 \times 10^{-2}\). Rounding to nearest integer implies x = 58.