The emf of the cell `Cr(s) abs(Cr^(3+)(1.0M))abs(Co^(2+)(1.0M))Co(s) [E^(o)" For " Cr^(3+)abs(Cr(s)=-0.74V & Co^(2+))Co(s) -0.28V]:`

Options

A. `-0.74 - (-0.28)=-0.46V`
B. `-0.74 + (-0.28)=-1.02V`
C. `-0.28-(-0.74)=+0.46V`
D. `0.74+0.28=+1.02 V`

Answer

To calculate the EMF (electromotive force) of the given electrochemical cell, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Anode and Cathode:**
- In the given cell representation `Cr(s) | Cr^(3+)(1.0M) | Co^(2+)(1.0M) | Co(s)`, chromium (Cr) is the anode and cobalt (Co) is the cathode.
- The anode is where oxidation occurs, and the cathode is where reduction occurs.

2. **Write the Half-Reactions:**
- For the anode (oxidation of chromium):
\[ \text{Cr}^{3+} + 3e^- \rightarrow \text{Cr(s)} \quad (E^\circ = -0.74 \, \text{V}) \]
- For the cathode (reduction of cobalt):
\[ \text{Co}^{2+} + 2e^- \rightarrow \text{Co(s)} \quad (E^\circ = -0.28 \, \text{V}) \]

3. **Determine the Standard Reduction Potentials:**
- Standard reduction potential for the cathode (Co): \( E^\circ_{\text{cathode}} = -0.28 \, \text{V} \)
- Standard reduction potential for the anode (Cr): \( E^\circ_{\text{anode}} = -0.74 \, \text{V} \)

4. **Calculate the EMF of the Cell:**
- The formula for calculating the EMF of the cell is:
\[ E_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
- Substituting the values:
\[ E_{\text{cell}} = (-0.28 \, \text{V}) - (-0.74 \, \text{V}) \]
- This simplifies to:
\[ E_{\text{cell}} = -0.28 + 0.74 = 0.46 \, \text{V} \]

5. **Final Answer:**
- The EMF of the cell is \( 0.46 \, \text{V} \).