The standard cell potential for the cell `Zn|Zn^(2+) (1 M)|| Cu^(2+) (1 M)| Cu`
Given `E_(Cu^(2+)//Cu)^(@) = 0.34 V` and `E_(Zn^(2+)//Zn)^(@) = - 0.76 V` is

Options

A. 1.10 V
B. ` - 0.42 V `
C. `-1.10 V`
D. None of these

Answer

To find the standard cell potential for the electrochemical cell `Zn|Zn^(2+) (1 M)|| Cu^(2+) (1 M)| Cu`, we can follow these steps:

### Step 1: Identify the half-reactions
- The half-reaction for copper is:
\[ Cu^{2+} + 2e^- \rightarrow Cu \quad (E^\circ = 0.34 \, V) \]
- The half-reaction for zinc is:
\[ Zn^{2+} + 2e^- \rightarrow Zn \quad (E^\circ = -0.76 \, V) \]

### Step 2: Determine the anode and cathode
- The anode is where oxidation occurs, and the cathode is where reduction occurs.
- Since zinc has a more negative standard reduction potential than copper, zinc will be oxidized (anode), and copper will be reduced (cathode).
- Thus, we have:
- Anode: \( Zn \rightarrow Zn^{2+} + 2e^- \)
- Cathode: \( Cu^{2+} + 2e^- \rightarrow Cu \)

### Step 3: Write the standard cell potential formula
The standard cell potential (E°cell) can be calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]

### Step 4: Substitute the values
Substituting the values we have:
\[ E^\circ_{cell} = E^\circ_{Cu^{2+}/Cu} - E^\circ_{Zn^{2+}/Zn} \]
\[ E^\circ_{cell} = 0.34 \, V - (-0.76 \, V) \]

### Step 5: Perform the calculation
Calculating the above expression:
\[ E^\circ_{cell} = 0.34 \, V + 0.76 \, V = 1.10 \, V \]

### Final Answer
The standard cell potential for the cell `Zn|Zn^(2+) (1 M)|| Cu^(2+) (1 M)| Cu` is:
\[ \boxed{1.10 \, V} \]