Consider the following `E^(@)` values: `E_(Li^(+)|Li)^(@)=-3.05V, E_(Cu^(2+)|Cu)^(@)=+0.34V`
Under similar conditions, the potential for the reaction `Cu+2Li^(+)rarrCu^(2+)+2Li`, is

Options

A. `-3.39V`
B. `+3.39V`
C. `-2.69V`
D. `+2.69V`

Answer

To find the potential for the reaction \( \text{Cu} + 2\text{Li}^+ \rightarrow \text{Cu}^{2+} + 2\text{Li} \), we can follow these steps:

### Step 1: Identify the half-reactions
The overall reaction can be broken down into two half-reactions:
1. The oxidation half-reaction: \( \text{Cu} \rightarrow \text{Cu}^{2+} + 2e^- \)
2. The reduction half-reaction: \( \text{Li}^+ + e^- \rightarrow \text{Li} \)

### Step 2: Write the standard reduction potentials
From the question, we have the standard reduction potentials:
- For \( \text{Li}^+ + e^- \rightarrow \text{Li} \): \( E^\circ = -3.05 \, \text{V} \)
- For \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \): \( E^\circ = +0.34 \, \text{V} \)

### Step 3: Convert the oxidation half-reaction to a reduction potential
The oxidation half-reaction for copper can be written as:
\[ \text{Cu} \rightarrow \text{Cu}^{2+} + 2e^- \]
To find the standard potential for the oxidation reaction, we take the negative of the reduction potential:
\[ E^\circ_{\text{oxidation}} = -E^\circ_{\text{reduction}} = -0.34 \, \text{V} \]

### Step 4: Combine the half-reactions
Now we can combine the two half-reactions. The overall cell potential \( E^\circ_{\text{cell}} \) is given by:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} + E^\circ_{\text{oxidation}} \]
Substituting the values we have:
\[ E^\circ_{\text{cell}} = (-3.05 \, \text{V}) + (-0.34 \, \text{V}) \]

### Step 5: Calculate the overall cell potential
Now we perform the addition:
\[ E^\circ_{\text{cell}} = -3.05 \, \text{V} - 0.34 \, \text{V} = -3.39 \, \text{V} \]

### Final Answer
The potential for the reaction \( \text{Cu} + 2\text{Li}^+ \rightarrow \text{Cu}^{2+} + 2\text{Li} \) is \( -3.39 \, \text{V} \).

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