To calculate the free energy ($\Delta G$) of the reaction at 25°C, we can proceed as follows:

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Step 1: Identify the half-reactions and standard potentials

The battery components indicate the following half-reactions:

1. Oxygen reduction reaction at the cathode: $\text{O}_2(g) + 4H^+ + 4e^- \rightarrow 2H_2O$ Standard electrode potential ($E^\circ$): $+1.23 \, \text{V}$.

2. Silver oxidation reaction at the anode: $\text{Ag}^+ + e^- \rightarrow \text{Ag}(s)$ Standard electrode potential ($E^\circ$): $+0.80 \, \text{V}$.

The overall cell reaction combines these half-reactions: $2\text{Ag}(s) + 2H^+ + \text{O}_2(g) \rightarrow 2\text{Ag}^+ + 2H_2O(l)$

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Step 2: Calculate the cell potential

The standard cell potential is given by: $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ Substitute values: $E^\circ_{\text{cell}} = 1.23 \, \text{V} - 0.80 \, \text{V} = 0.43 \, \text{V}$

Now, incorporate the Nernst equation to account for non-standard conditions.

The Nernst equation:

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

• $R = 8.314 \, \text{J/mol·K}$, universal gas constant.
• $T = 298 \, \text{K}$, temperature.
• $n = 4$, number of electrons transferred.
• $F = 96485 \, \text{C/mol}$, Faraday's constant.
• $Q$, reaction quotient: [ Q = \frac{[\text{Ag}^+]^2}{[H^+]^2 \cdot P_{\text{O}_2}} ]

Concentrations and partial pressures:

• $[H^+] = 10^{-\text{pH}} = 10^{-2} \, \text{M}$,
• $[\text{Ag}^+] = 0.1 \, \text{M}$,
• $P_{\text{O}_2} = 1 \, \text{bar}$.

Thus: $Q = \frac{(0.1)^2}{(10^{-2})^2 \cdot 1} = \frac{0.01}{0.0001} = 100$

Substitute into the Nernst equation: $E_{\text{cell}} = 0.43 - \frac{8.314 \cdot 298}{4 \cdot 96485} \ln 100$

First, calculate the temperature term: $\frac{8.314 \cdot 298}{4 \cdot 96485} = 0.00643$

Now compute the logarithm term: $\ln 100 = 4.605$

Substitute back: $E_{\text{cell}} = 0.43 - 0.00643 \cdot 4.605$ $E_{\text{cell}} = 0.43 - 0.0296 = 0.4004 \, \text{V}$

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Step 3: Calculate the free energy

The relationship between free energy ($\Delta G$) and cell potential is: $\Delta G = -nFE_{\text{cell}}$

Substitute values: $\Delta G = -4 \cdot 96485 \cdot 0.4004$ $\Delta G = -154451 \, \text{J/mol} = -154.45 \, \text{kJ/mol}$

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Final Answer:

The free energy of the reaction at 25°C is: $\Delta G = -154.45 \, \text{kJ/mol}$

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Do you want further details or have any questions?
Here are 5 related questions to deepen your understanding:

1. How does the pH influence the reaction quotient ($Q$) and cell potential?
2. What happens to $\Delta G$ if the concentration of $\text{Ag}^+$ changes?
3. How does temperature affect the cell potential and $\Delta G$?
4. Can this battery produce electrical energy if $E_{\text{cell}} \leq 0$?
5. Why is $n = 4$ used in this calculation?

Tip: Always check if the reaction quotient ($Q$) is dimensionless and use consistent units when calculating $\Delta G$.