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 of the cell given by the equation:

\[ \text{Cr(s)} | \text{Cr}^{3+}(1.0M) | \text{Co}^{2+}(1.0M) | \text{Co(s)} \]

we will follow these steps:

### Step 1: Identify the Standard Reduction Potentials
We are given the standard reduction potentials:
- For the reaction \( \text{Cr}^{3+} + 3e^- \rightarrow \text{Cr(s)} \), \( E^\circ = -0.74 \, \text{V} \)
- For the reaction \( \text{Co}^{2+} + 2e^- \rightarrow \text{Co(s)} \), \( E^\circ = -0.28 \, \text{V} \)

### Step 2: Determine Anode and Cathode
The species with the higher reduction potential will act as the cathode (reduction occurs here), and the species with the lower reduction potential will act as the anode (oxidation occurs here).
- Since \( -0.28 \, \text{V} > -0.74 \, \text{V} \), cobalt will be the cathode and chromium will be the anode.

### Step 3: Calculate the Standard EMF of the Cell
The standard EMF of the cell (\( E^\circ_{\text{cell}} \)) can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Substituting the values:
\[ E^\circ_{\text{cell}} = (-0.28 \, \text{V}) - (-0.74 \, \text{V}) \]
\[ E^\circ_{\text{cell}} = -0.28 + 0.74 \]
\[ E^\circ_{\text{cell}} = 0.46 \, \text{V} \]

### Step 4: Write the Half-Reactions
- Anode (oxidation):
\[ \text{Cr(s)} \rightarrow \text{Cr}^{3+} + 3e^- \]
- Cathode (reduction):
\[ \text{Co}^{2+} + 2e^- \rightarrow \text{Co(s)} \]

### Step 5: Equalize the Number of Electrons
To combine the half-reactions, we need to equalize the number of electrons transferred. The least common multiple of 3 and 2 is 6. Therefore:
- Multiply the chromium half-reaction by 2:
\[ 2\text{Cr(s)} \rightarrow 2\text{Cr}^{3+} + 6e^- \]
- Multiply the cobalt half-reaction by 3:
\[ 3\text{Co}^{2+} + 6e^- \rightarrow 3\text{Co(s)} \]

### Step 6: Combine the Half-Reactions
Combining the two half-reactions gives:
\[ 2\text{Cr(s)} + 3\text{Co}^{2+} \rightarrow 2\text{Cr}^{3+} + 3\text{Co(s)} \]

### Step 7: Apply the Nernst Equation
Now we will use the Nernst equation to find the cell potential under non-standard conditions:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \frac{[\text{Cr}^{3+}]^2}{[\text{Co}^{2+}]^3} \]

Where:
- \( n = 6 \) (total number of electrons transferred)
- Both concentrations are 1.0 M, hence:
\[ E_{\text{cell}} = 0.46 \, \text{V} - \frac{0.0591}{6} \log \frac{1^2}{1^3} \]
\[ E_{\text{cell}} = 0.46 \, \text{V} - \frac{0.0591}{6} \log 1 \]
Since \( \log 1 = 0 \):
\[ E_{\text{cell}} = 0.46 \, \text{V} - 0 \]
\[ E_{\text{cell}} = 0.46 \, \text{V} \]

### Final Answer
The EMF of the cell is \( 0.46 \, \text{V} \).

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