How much electricity in terms of Faraday is required to produce $40.0$ g of Al from molten $ \mathrm{Al_2O_3} $?

Question

Consider following reaction :
$Fe(OH)_2 + 2e^- \rightarrow Fe(s) + 2OH^-$ , $E^{\circ} = -0.88 \text{ V}$
$AgBr(s) + e^- \rightarrow Ag(s) + Br^-(aq)$ , $E^{\circ} = 0.07 \text{ V}$
Select the correct statement.
(1) $E^{\circ}_{cell} = -0.95 \text{V}$
(2) $E^{\circ}_{cell}$ is an extensive property
(3) $Fe$ is getting reduced
(4) Net cell reaction: $Fe(s) + 2OH^-(aq) + 2AgBr(s) \rightarrow Fe(OH)_2 + 2Ag(s) + 2Br^-(aq)$

Answer 1

Step 1: Write the electrode reaction for aluminium.
In molten $ \mathrm{Al_2O_3} $, aluminium ion is reduced at the cathode as follows:
\[ \mathrm{Al^{3+} + 3e^- \rightarrow Al} \] This shows that 1 mole of aluminium requires 3 moles of electrons, that is, 3 Faradays of electricity.
Step 2: Calculate the number of moles of aluminium.
Molar mass of aluminium $= 27\ \mathrm{g\ mol^{-1}}$
Given mass of aluminium $= 40.0\ \mathrm{g}$
\[ \text{Moles of Al} = \frac{40.0}{27} \] \[ = 1.4815\ \text{mol} \] Step 3: Calculate Faradays required.
Since $1$ mole of Al requires $3$ F,
\[ 1.4815 \times 3 = 4.4445\ \mathrm{F} \] \[ \approx 4.44\ \mathrm{F} \] Step 4: Conclusion.
Therefore, the amount of electricity required to produce $40.0$ g of aluminium is 4.44 F.
Final Answer:4.44 F.

How much electricity in terms of Faraday is required to produce \(40.0\) g of Al from molten \( \mathrm{Al_2O_3} \)?

Show Hint

The Correct Option is A

Solution and Explanation

Top Questions on Electrochemistry

Questions Asked in GUJCET exam