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Step-by-Step Solution
Step 1: Identify the Given Data
• Current, I = 2\,\text{A}
• Time, t = 8\,\text{minutes} = 8 \times 60 = 480 \,\text{s}
• Faraday constant, F = 96000\,\text{C mol}^{-1}
• Atomic mass of chromium = 52
• Mass of \text{Cr}^{3+} actually obtained = 0.104\,\text{g}
Step 2: Write the Relevant Half-Reaction
The reduction of dichromate ions in acidic medium is given by:
Cr_2O_7^{2-} + 14\,H^+ + 6\,e^- \;\longrightarrow\; 2\,Cr^{3+} + 7\,H_2O
From this half-reaction, note that 6\,\text{electrons} reduce 2\,\text{Cr}^{3+} ions. Therefore, for each Cr^{3+} ion formed, 3\,\text{electrons} are required.
Step 3: Calculate the Theoretical Mass of \text{Cr}^{3+} Using Faraday's Laws
According to Faraday's first law, the mass of a substance deposited or liberated (here, w_{Cr^{3+}} ) is given by:
w_{Cr^{3+}} = \frac{E \times I \times t}{F}
where E is the equivalent mass. For chromium going from +6 to +3 , each chromium requires 3 electrons, so the equivalent mass of chromium is:
E = \frac{\text{Atomic mass of Cr}}{\text{Number of electrons gained per Cr}}
= \frac{52}{3}.
Substituting the values,
w_{Cr^{3+}}
= \frac{\left(\tfrac{52}{3}\right) \times 2 \,\text{A} \times 480 \,\text{s}}{96000\ \text{C mol}^{-1}}
Simplify step by step:
w_{Cr^{3+}}
= \frac{52}{3} \times \frac{2 \times 480}{96000}
= \frac{52}{3} \times \frac{960}{96000}
= \frac{52}{3} \times \frac{1}{100}
= \frac{52}{300}
\approx 0.1733\,\text{g}.
So, the theoretical mass of Cr^{3+} that should be obtained is 0.1733\,\text{g} .
Step 4: Compute the Percentage Efficiency
We compare the actual mass obtained ( 0.104\,\text{g} ) to the theoretical mass ( 0.1733\,\text{g} ):
\%\ \text{efficiency}
= \frac{\text{Actual mass}}{\text{Theoretical mass}} \times 100
= \frac{0.104}{0.1733} \times 100
\approx 60\%.
Final Answer
The efficiency of the process is approximately 60\% .
