One plate of Cu and one plate of Ag are placed in a solution containing cupric and silver ions. What reactions would occur if the concentration of each species were 1 M ? Given that reduction potential of `Cu^(2+)//Cu and Ag^(+)//Ag` electrodes are +0.34 and + 0.80 volt respectively.

Explanation

To solve the problem, we need to analyze the electrochemical reactions that occur when a copper (Cu) plate and a silver (Ag) plate are placed in a solution containing cupric ions (Cu²⁺) and silver ions (Ag⁺), both at a concentration of 1 M. We will also use the given standard reduction potentials for the half-reactions.

### Step-by-Step Solution:

1. **Identify the half-reactions and their standard reduction potentials:**
- The reduction half-reaction for copper is:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad (E^\circ = +0.34 \, \text{V}) \]
- The reduction half-reaction for silver is:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \quad (E^\circ = +0.80 \, \text{V}) \]

2. **Determine which metal will act as the anode and which as the cathode:**
- The metal with the higher reduction potential will be reduced (cathode), and the metal with the lower reduction potential will be oxidized (anode).
- Here, Ag has a higher reduction potential (+0.80 V) compared to Cu (+0.34 V), so:
- **Ag will be the cathode** (reduction occurs here).
- **Cu will be the anode** (oxidation occurs here).

3. **Write the oxidation and reduction reactions:**
- **Oxidation at the anode (Cu):**
\[ \text{Cu} \rightarrow \text{Cu}^{2+} + 2e^- \]
- **Reduction at the cathode (Ag):**
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]

4. **Balance the overall reaction:**
- To balance the electrons, we need to multiply the silver reduction reaction by 2:
\[ 2 \text{Ag}^+ + 2e^- \rightarrow 2 \text{Ag} \]
- Now, combine the oxidation and reduction reactions:
\[ \text{Cu} + 2 \text{Ag}^+ \rightarrow \text{Cu}^{2+} + 2 \text{Ag} \]

5. **Calculate the standard cell potential (E°cell):**
- Using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
- Substitute the values:
\[ E^\circ_{\text{cell}} = 0.80 \, \text{V} - 0.34 \, \text{V} = 0.46 \, \text{V} \]
- Since \(E^\circ_{\text{cell}}\) is positive, the reaction is feasible.

6. **Conclusion:**
- The overall reaction that occurs is:
\[ \text{Cu} + 2 \text{Ag}^+ \rightarrow \text{Cu}^{2+} + 2 \text{Ag} \]
- This indicates that copper will oxidize to copper ions, while silver ions will reduce to solid silver.