To solve the problem of calculating the percentage efficiency of the electrochemical cell, we will follow these steps:
### Step 1: Calculate the total charge passed through the cell
The total charge (Q) can be calculated using the formula:
\[ Q = I \times t \]
where:
- \( I = 2 \, \text{A} \) (current)
- \( t = 5 \, \text{min} = 5 \times 60 \, \text{s} = 300 \, \text{s} \)
Calculating the charge:
\[ Q = 2 \, \text{A} \times 300 \, \text{s} = 600 \, \text{C} \]
### Step 2: Convert the charge to moles of electrons
Using Faraday's constant, which is approximately \( 96500 \, \text{C/mol} \):
\[ \text{Moles of electrons} = \frac{Q}{F} = \frac{600 \, \text{C}}{96500 \, \text{C/mol}} \approx 0.0062 \, \text{mol} \]
### Step 3: Determine the moles of \( Cr^{3+} \) produced
From the balanced equation:
\[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]
We see that 6 moles of electrons produce 2 moles of \( Cr^{3+} \). Therefore, the moles of \( Cr^{3+} \) produced can be calculated as:
\[ \text{Moles of } Cr^{3+} = \frac{2}{6} \times \text{Moles of electrons} = \frac{1}{3} \times 0.0062 \, \text{mol} \approx 0.00207 \, \text{mol} \]
### Step 4: Calculate the theoretical mass of \( Cr^{3+} \)
Using the atomic mass of chromium (Cr = 52 g/mol):
\[ \text{Theoretical mass} = \text{Moles of } Cr^{3+} \times \text{Atomic mass} = 0.00207 \, \text{mol} \times 52 \, \text{g/mol} \approx 0.10764 \, \text{g} \]
### Step 5: Calculate the percentage efficiency of the cell
The percentage efficiency can be calculated using the formula:
\[ \text{Percentage efficiency} = \left( \frac{\text{Actual mass}}{\text{Theoretical mass}} \right) \times 100 \]
Given that the actual mass of \( Cr^{3+} \) produced is \( 0.104 \, \text{g} \):
\[ \text{Percentage efficiency} = \left( \frac{0.104 \, \text{g}}{0.10764 \, \text{g}} \right) \times 100 \approx 96.50\% \]
### Final Answer:
The percentage efficiency of the cell is approximately **96.50%**.
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