Consider the following reaction,
`Cu|Cu^(2+) (1M)||(Zn^(2+)(1 M)|Zn`
A cell represented above should have emf

Options

A. positive
B. negative
C. zero
D. Cannot be predicted

Explanation

To determine the EMF (Electromotive Force) of the electrochemical cell represented by the notation `Cu|Cu^(2+) (1M)||(Zn^(2+)(1 M)|Zn`, we will follow these steps:

### Step 1: Identify the Half-Reactions
In the given cell, we have two half-reactions:
1. **Reduction at the Cathode (Copper)**:
\[ Cu^{2+} + 2e^- \rightarrow Cu \quad (E^\circ_{Cu} = +0.34 \, V) \]
2. **Oxidation at the Anode (Zinc)**:
\[ Zn \rightarrow Zn^{2+} + 2e^- \quad (E^\circ_{Zn} = -0.76 \, V) \]

### Step 2: Write the Cell Notation
The cell notation indicates that copper is being reduced (gaining electrons) and zinc is being oxidized (losing electrons).

### Step 3: Calculate the Standard EMF of the Cell
The standard EMF (E°) of the cell can be calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
Substituting the values:
\[ E^\circ_{cell} = E^\circ_{Cu} - E^\circ_{Zn} = 0.34 \, V - (-0.76 \, V) \]
\[ E^\circ_{cell} = 0.34 \, V + 0.76 \, V = 1.10 \, V \]

### Step 4: Apply the Nernst Equation
The Nernst equation is used to calculate the EMF under non-standard conditions:
\[ E = E^\circ - \frac{0.059}{n} \log \left( \frac{[products]}{[reactants]} \right) \]
Where:
- \( n \) = number of moles of electrons transferred (which is 2 in this case).
- The concentrations of both \( Cu^{2+} \) and \( Zn^{2+} \) are given as 1 M.

### Step 5: Substitute Values into the Nernst Equation
Since both concentrations are 1 M, the reaction quotient \( Q \) becomes:
\[ Q = \frac{[Cu^{2+}]}{[Zn^{2+}]} = \frac{1}{1} = 1 \]
Thus, \( \log(1) = 0 \).

Substituting into the Nernst equation:
\[ E = 1.10 \, V - \frac{0.059}{2} \cdot 0 \]
\[ E = 1.10 \, V \]

### Conclusion
The EMF of the cell is:
\[ \text{EMF} = 1.10 \, V \]