Calculate the potential of the following cell `:`
`Pt|underset((2.0M)(1.0M))(Co^(2+),Co^(3+))|underset((1.0M)(4.0M)(1.0M))(Cr^(3+),Cr_(2)O_(7)^(2_-),H^(o+))|Pt`
`E^(c-)._(Co^(2+)|Co^(3+))=-2V,E^(c-)._(Cr_(2)O_(7)^(2-)|Cr^(3+))=+1.0V`

Options

A. `1.024V`
B. `-1.024V`
C. `0.976V`
D. `-0.976V`

Explanation

To calculate the potential of the given electrochemical cell, we will follow these steps:

### Step 1: Identify the half-reactions and their standard electrode potentials (E°)

We have the following half-reactions and their standard potentials:
1. \( \text{Co}^{2+} + 1e^- \rightarrow \text{Co}^{3+} \) with \( E^\circ = -2 \, \text{V} \)
2. \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \) with \( E^\circ = +1 \, \text{V} \)

### Step 2: Write the oxidation and reduction reactions

- **Anode (oxidation)**: \( \text{Co}^{2+} \rightarrow \text{Co}^{3+} + e^- \)
- **Cathode (reduction)**: \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \)

### Step 3: Balance the half-reactions

To balance the half-reactions:
- The oxidation reaction involves 1 electron.
- The reduction reaction involves 6 electrons.

We need to multiply the oxidation half-reaction by 6 to balance the electrons:
\[ 6\text{Co}^{2+} \rightarrow 6\text{Co}^{3+} + 6e^- \]

### Step 4: Combine the half-reactions

Now, we can combine the balanced half-reactions:
\[ 6\text{Co}^{2+} + \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \rightarrow 6\text{Co}^{3+} + 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]

### Step 5: Calculate the standard cell potential (E°cell)

Using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
\[ E^\circ_{\text{cell}} = (+1 \, \text{V}) - (-2 \, \text{V}) = 1 + 2 = 3 \, \text{V} \]

### Step 6: Calculate the reaction quotient (Q)

Using the concentrations given:
- \( [\text{Co}^{2+}] = 2.0 \, \text{M} \)
- \( [\text{Co}^{3+}] = 1.0 \, \text{M} \)
- \( [\text{Cr}^{3+}] = 1.0 \, \text{M} \)
- \( [\text{Cr}_2\text{O}_7^{2-}] = 4.0 \, \text{M} \)
- \( [\text{H}^+] = 1.0 \, \text{M} \)

The reaction quotient \( Q \) is given by:
\[ Q = \frac{[\text{Co}^{3+}]^6 [\text{Cr}^{3+}]^2}{[\text{Co}^{2+}]^6 [\text{Cr}_2\text{O}_7^{2-}] [\text{H}^+]^{14}} \]
Substituting the values:
\[ Q = \frac{(1)^6 (1)^2}{(2)^6 (4)(1)^{14}} = \frac{1}{64 \cdot 4} = \frac{1}{256} \]

### Step 7: Calculate the cell potential (Ecell) using the Nernst equation

Using the Nernst equation:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q \]
Where \( n = 6 \) (number of electrons transferred):
\[ E_{\text{cell}} = 3 - \frac{0.0591}{6} \log \left(\frac{1}{256}\right) \]

Calculating \( \log(256) \):
\[ \log(256) = 2.408 \]
So,
\[ E_{\text{cell}} = 3 - \frac{0.0591}{6} \cdot (-2.408) \]
\[ E_{\text{cell}} = 3 + 0.02365 \cdot 2.408 \]
\[ E_{\text{cell}} = 3 + 0.0569 \]
\[ E_{\text{cell}} \approx 3.0569 \, \text{V} \]

### Final Answer
The potential of the cell is approximately \( 3.0569 \, \text{V} \).

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