Question

The cell potential at 25 °c with concentrations [co₃] = 0.706 m, [co₂] = 0.400 m, [cl⁻] = 0.390 m, and cl₂ pressure (pcl₂) = 8.80 atm is determined by the nernst equation and is calculated as ecell = 0.45 v.

a. \( E_{\text{cell}} = 0.45 \, V + \frac{0.0592}{n} \log \left( \frac{[CO₂]}{[CO₃]} \right) + \frac{0.0592}{n} \log \left( \frac{[\text{Cl}^-]}{P_{\text{Cl}_2}} \right) \)
b. \( E_{\text{cell}} = 0.45 \, V - \frac{0.0592}{n} \log \left( \frac{[CO₃]}{[CO₂]} \right) + \frac{0.0592}{n} \log \left( \frac{[\text{Cl}^-]}{P_{\text{Cl}_2}} \right) \)
c. \( E_{\text{cell}} = 0.45 \, V + \frac{0.0592}{n} \log \left( \frac{[CO₃]}{[CO₂]} \right) - \frac{0.0592}{n} \log \left( \frac{[\text{Cl}^-]}{P_{\text{Cl}_2}} \right) \)
d. \( E_{\text{cell}} = 0.45 \, V - \frac{0.0592}{n} \log \left( \frac{[CO₂]}{[CO₃]} \right) - \frac{0.0592}{n} \log \left( \frac{[\text{Cl}^-]}{P_{\text{Cl}_2}} \right) \)

Answer 1

The correct answer is option (a):
`\[ E_{\text{cell}} = 0.45 \, V + (0.0592)/(n) \log \left( ([CO₂])/([CO₃]) \right) + (0.0592)/(n) \log \left( \frac{[\text{Cl}^-]}{P_{\text{Cl}_2}} \right) \]`

What is Nernst equation?

The Nernst equation is given by:

`\[ E_{\text{cell}} = E_{\text{standard}} - (0.0592)/(n) \log \left( (Q)/(P) \right) \]`

where:

`\( E_{\text{cell}} \)` = cell potential,

`\( E_{\text{standard}} \)` = standard cell potential,

n = number of moles of electrons exchanged in the balanced chemical equation,

Q = reaction quotient,

P = pressure of any gas involved in the reaction.

Comparing this with the given equation
`\( E_{\text{cell}} = 0.45 \, V + (0.0592)/(n) \log \left( ([CO₂])/([CO₃]) \right) + (0.0592)/(n) \log \left( \frac{[\text{Cl}^-]}{P_{\text{Cl}_2}} \right) \)`, see that the correct representation is:

`\[ E_{\text{cell}} = 0.45 \, V + (0.0592)/(n) \log \left( ([CO₂])/([CO₃]) \right) + (0.0592)/(n) \log \left( \frac{[\text{Cl}^-]}{P_{\text{Cl}_2}} \right) \]`

So, the correct answer is option (a):

`\[ E_{\text{cell}} = 0.45 \, V + (0.0592)/(n) \log \left( ([CO₂])/([CO₃]) \right) + (0.0592)/(n) \log \left( \frac{[\text{Cl}^-]}{P_{\text{Cl}_2}} \right) \]`