E.m.f of the cell
`2Ag^(+) + Cu rightarrow Cu^(+2) = 2Ag`
[Given : `E^(@)`= -0.8V , `E_(Cu^(+2))^@` Cu = 0.3V]`

Options

A. `-0.5V`
B. `0.5V`
C. `-1.1V`
D. `1.1V`

Answer

To calculate the electromotive force (E.M.F) of the cell reaction given by:

\[ 2Ag^{+} + Cu \rightarrow Cu^{2+} + 2Ag \]

we will follow these steps:

### Step 1: Identify the Half-Reactions
The overall cell reaction can be separated into two half-reactions:
- **Reduction half-reaction** (occurs at the cathode):
\[ 2Ag^{+} + 2e^{-} \rightarrow 2Ag \]
- **Oxidation half-reaction** (occurs at the anode):
\[ Cu \rightarrow Cu^{2+} + 2e^{-} \]

### Step 2: Write Down the Standard Electrode Potentials
From the problem, we have the following standard electrode potentials:
- For the reduction of silver:
\[ E^{\circ}(Ag^{+}/Ag) = -0.8 \, V \]
- For the reduction of copper:
\[ E^{\circ}(Cu^{2+}/Cu) = 0.3 \, V \]

### Step 3: Determine the E.M.F of the Cell
The E.M.F of the cell (E_cell) can be calculated using the formula:
\[ E_{cell} = E_{cathode} - E_{anode} \]

Here:
- The cathode is where reduction occurs (Ag):
\[ E_{cathode} = E^{\circ}(Ag^{+}/Ag) = -0.8 \, V \]
- The anode is where oxidation occurs (Cu):
\[ E_{anode} = E^{\circ}(Cu^{2+}/Cu) = 0.3 \, V \]

### Step 4: Substitute the Values into the Formula
Substituting the values into the formula gives:
\[ E_{cell} = (-0.8 \, V) - (0.3 \, V) \]
\[ E_{cell} = -0.8 \, V - 0.3 \, V \]
\[ E_{cell} = -1.1 \, V \]

### Step 5: Correct the Sign
Since we are looking for the E.M.F of the cell, we need to take the absolute value:
\[ E_{cell} = 1.1 \, V \]

### Final Answer
The E.M.F of the cell is:
\[ E_{cell} = 1.1 \, V \]

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