Question

Molten \(Al_2O_3\) was electrolyzed between carbon electrodes. The mass (in g) of aluminium produced at cathode when 965 amperes current is passed through it for 1000 seconds is (\(F=96500\,C\,mol^{-1}\))

• A. 30
• B. 90
• C. 60
• D. 45

Hint:

For electrolysis problems: \[ m=\frac{MIt}{nF} \] Always determine \(n\) from the cathode half-reaction before substituting values.

Answer 1

The Correct Option is B

Step 1: Recall Faraday's first law.
The mass deposited in electrolysis is \[ m = \frac{M\,I\,t}{n\,F}, \] where $M$ is molar mass, $I$ is current, $t$ is time, $n$ is electrons per ion, and $F$ is the Faraday constant.
Step 2: Write the cathode reaction.
Aluminium ion gains three electrons: \[ Al^{3+} + 3e^- \rightarrow Al. \] So $n = 3$.
Step 3: Find the charge passed.
\[ Q = I\,t = 965 \times 1000 = 965000 \ \text{C}. \]
Step 4: Note the molar mass.
The molar mass of aluminium is $M = 27 \ \text{g mol}^{-1}$.
Step 5: Plug into the formula.
\[ m = \frac{27 \times 965000}{3 \times 96500} = \frac{27 \times 10}{3}. \] Here $\dfrac{965000}{96500} = 10$.
Step 6: Compute and conclude.
\[ m = \frac{270}{3} = 90 \ \text{g}. \] So the aluminium deposited is \[ \boxed{90 \ \text{g}} \]