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Step-by-Step Solution
Step 1: Recall the Nernst Equation at Equilibrium
The Nernst equation for a galvanic cell is given by:
$ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \dfrac{2.303\,R\,T}{n\,F}\,\log K $
At equilibrium, the cell potential $E_{\text{cell}}$ becomes 0, so we set $E_{\text{cell}} = 0$.
Step 2: Substitute the Given Data
Given:
Standard cell potential, $E^{\circ}_{\text{cell}} = 0.295\,\text{V}$
Number of electrons, $n = 2$
Temperature, $T = 298\,\text{K}$
Gas constant, $R = 8.314\,\text{J\,mol}^{-1}\text{K}^{-1}$
Faraday constant, $F = 96500\,\text{C\,mol}^{-1}$
At $25^{\circ}\text{C}$ ($298\,\text{K}$), the factor $ \dfrac{2.303\,R\,T}{F} $ is approximately $0.0591\,\text{V} $.
Step 3: Set Up the Equation at Equilibrium
Since $E_{\text{cell}} = 0$ at equilibrium:
$ 0 = E^{\circ}_{\text{cell}} - \dfrac{2.303\,R\,T}{n\,F}\,\log K $
Rearranging:
$ E^{\circ}_{\text{cell}} = \dfrac{2.303\,R\,T}{n\,F}\,\log K $
We can substitute the known values:
$ 0.295 = \left(\dfrac{0.0591}{2}\right) \log K
$
Step 4: Solve for the Equilibrium Constant K
First simplify the right-hand side:
$ \dfrac{0.0591}{2} = 0.02955
$
Hence, we get:
$ 0.295 = 0.02955 \,\log K
$
Divide both sides by 0.02955:
$ \log K = \dfrac{0.295}{0.02955} = 10
$
Therefore:
$ K = 10^{10}
$
Final Answer
The equilibrium constant of the reaction is $ 1.0 \times 10^{10} $.
