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For the following cell with gas electrod...

For the following cell with gas electrodes at `p_(1) and p_(2)` as shown:
`underset(at P1)(PtCl_(2))underset(1M)(abs(HCl)) underset(at P2)(Pt (Cl_(2))` cell reaction is spontaneous if :

A

`P_(1)=P_(2)`

B

`P_(1) gt P_(2)`

C

`P_(1) lt P_(2)`

D

`P_(2) = 0`

Text Solution

AI Generated Solution

The correct Answer is:
To determine the conditions under which the cell reaction is spontaneous, we can follow these steps: ### Step 1: Identify the Cell Components We have a cell with gas electrodes at pressures \( P_1 \) and \( P_2 \). The components are: - At \( P_1 \): \( \text{Pt} | \text{Cl}_2 | \text{HCl (1M)} \) - At \( P_2 \): \( \text{Pt} | \text{Cl}_2 \) ### Step 2: Determine the Half-Reactions - **At the Cathode (Reduction)**: Chlorine gas (\( \text{Cl}_2 \)) is reduced to chloride ions (\( \text{Cl}^- \)): \[ \text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^- \] - **At the Anode (Oxidation)**: Chloride ions (\( \text{Cl}^- \)) are oxidized to chlorine gas (\( \text{Cl}_2 \)): \[ 2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^- \] ### Step 3: Write the Nernst Equation The Nernst equation for the cell can be expressed as: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q \] Where: - \( n = 2 \) (number of electrons transferred) - \( Q \) is the reaction quotient. For our cell, since it is a concentration cell, \( E^\circ_{\text{cell}} = 0 \). Thus, the equation simplifies to: \[ E_{\text{cell}} = - \frac{0.0591}{2} \log \left( \frac{P_1}{P_2} \right) \] ### Step 4: Rearranging the Equation We can rearrange the equation to express it in terms of \( P_2 \) and \( P_1 \): \[ E_{\text{cell}} = \frac{0.0591}{2} \log \left( \frac{P_2}{P_1} \right) \] ### Step 5: Condition for Spontaneity For the cell reaction to be spontaneous, \( E_{\text{cell}} \) must be positive: \[ \frac{0.0591}{2} \log \left( \frac{P_2}{P_1} \right) > 0 \] This implies: \[ \log \left( \frac{P_2}{P_1} \right) > 0 \] Which means: \[ \frac{P_2}{P_1} > 1 \quad \Rightarrow \quad P_2 > P_1 \] ### Conclusion The cell reaction is spontaneous if \( P_2 > P_1 \). ### Final Answer The correct condition for spontaneity is when \( P_2 \) is greater than \( P_1 \). ---
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