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 (\(F=96500\,C\,mol^{-1}\))
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 (\(F=96500\,C\,mol^{-1}\))
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For electrolysis problems:
\[
m=\frac{MIt}{nF}
\]
Always determine \(n\) from the cathode half-reaction before substituting values.
Updated On: Jun 17, 2026
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The Correct Option is B
Solution and Explanation
Concept:
According to Faraday's First Law,
\[
m=\frac{MIt}{nF}
\]
where
• \(m\) = mass deposited
• \(M\) = molar mass
• \(I\) = current
• \(t\) = time
• \(n\) = electrons involved
• \(F\) = Faraday constant
Step 1: Write cathode reaction. \[ Al^{3+}+3e^- \rightarrow Al \] Thus, \[ n=3 \]
Step 2: Calculate charge passed. \[ Q=It \] \[ =(965)(1000) \] \[ =965000C \]
Step 3: Calculate mass deposited. Molar mass of aluminium: \[ M=27 \] \[ m= \frac{27\times965000} {3\times96500} \] \[ = \frac{27\times10}{3} \] \[ =90g \]
Step 4: Final conclusion. \[ \boxed{90g} \] Hence, \[ \boxed{\text{Option (B)}} \]
• \(m\) = mass deposited
• \(M\) = molar mass
• \(I\) = current
• \(t\) = time
• \(n\) = electrons involved
• \(F\) = Faraday constant
Step 1: Write cathode reaction. \[ Al^{3+}+3e^- \rightarrow Al \] Thus, \[ n=3 \]
Step 2: Calculate charge passed. \[ Q=It \] \[ =(965)(1000) \] \[ =965000C \]
Step 3: Calculate mass deposited. Molar mass of aluminium: \[ M=27 \] \[ m= \frac{27\times965000} {3\times96500} \] \[ = \frac{27\times10}{3} \] \[ =90g \]
Step 4: Final conclusion. \[ \boxed{90g} \] Hence, \[ \boxed{\text{Option (B)}} \]
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