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2Ag^(+)(aq) + Cu(s) rightarrow Cu^(2+)(a...

`2Ag^(+)(aq) + Cu(s) rightarrow Cu^(2+)(aq) + 2Ag(s)`
The standard potential for this reaction is 0.46 V. Which change will increase the potential the most?

A

Doubling the `[Ag^+]`

B

Halving the `[Cu^(2+)]`

C

Doubling the size of the Cu(s) electrode

D

Decreasing the size of the Ag electrode by one - half

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the given electrochemical reaction and understand how changes in concentration affect the cell potential according to the Nernst equation. ### Step-by-Step Solution: 1. **Identify the Reaction and Standard Potential:** The given reaction is: \[ 2Ag^+(aq) + Cu(s) \rightarrow Cu^{2+}(aq) + 2Ag(s) \] The standard potential (\(E^0\)) for this reaction is given as 0.46 V. 2. **Write the Nernst Equation:** The Nernst equation relates the cell potential (\(E\)) to the standard potential (\(E^0\)) and the concentrations of the reactants and products: \[ E = E^0 - \frac{0.059}{n} \log \left( \frac{[Cu^{2+}]}{[Ag^+]^2} \right) \] Here, \(n\) is the number of moles of electrons transferred in the reaction, which is 2 for this reaction. 3. **Analyze the Effect of Concentration Changes:** The term \(\frac{[Cu^{2+}]}{[Ag^+]^2}\) is crucial in determining the cell potential. To increase the cell potential \(E\), we need to minimize the value of the logarithmic term. - **Doubling the Concentration of \(Ag^+\):** If we double the concentration of \(Ag^+\), the term becomes: \[ \frac{[Cu^{2+}]}{(2[Ag^+])^2} = \frac{[Cu^{2+}]}{4[Ag^+]^2} \] This increases the denominator, thus decreasing the value of the logarithmic term, which leads to an increase in \(E\). - **Halving the Concentration of \(Cu^{2+}\):** If we halve the concentration of \(Cu^{2+}\), the term becomes: \[ \frac{(0.5[Cu^{2+}])}{[Ag^+]^2} \] This increases the value of the logarithmic term, leading to a decrease in \(E\). - **Changing the Size of Electrodes:** Changing the size of the electrodes does not affect the concentrations of the ions in solution, hence it does not affect the potential. 4. **Conclusion:** The change that will increase the potential the most is **doubling the concentration of \(Ag^+\)**, as it significantly reduces the logarithmic term in the Nernst equation. ### Final Answer: The change that will increase the potential the most is **doubling the concentration of \(Ag^+\)**. ---
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