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: Applying Nernst Equation.
The Nernst equation for a cell is given by: \[ E_{\text{cell}} = E^o_{\text{cell}} - \frac{0.06}{n} \log Q \] Where: - \( E_{\text{cell}} \) is the cell potential, - \( E^o_{\text{cell}} \) is the standard cell potential, - \( n \) is the number of moles of electrons involved in the reaction, - \( Q \) is the reaction quotient.
Step 2: Calculate the standard cell potential.
We can calculate \( E^o_{\text{cell}} \) using the given standard electrode potentials for the two half-reactions: \[ E^o_{\text{cell}} = E^o_{\text{Bi$_2$O$_3$ / Bi}} - E^o_{\text{[Sn(OH)$_3$]}} \] Substituting the given values: \[ E^o_{\text{cell}} = -0.44 - (-0.90) = +0.46 \, \text{V} \]
Step 3: Apply the Nernst equation.
Now, applying the Nernst equation: \[ E_{\text{cell}} = 0.46 - \frac{0.06}{6} \log 10^6 \] \[ E_{\text{cell}} = 0.46 - \frac{0.06}{6} \times 6 \] \[ E_{\text{cell}} = 0.46 - 0.06 = 0.46 \, \text{V} \]
Step 4: Conclusion.
We are given that \( E_{\text{cell}} = x \times 10^{-1} \). Therefore, \[ x = 4 \] Thus, the correct answer is (4).