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Faraday's laws
Concepts tested here
- Charge for oxidation
- Faraday's second law
All Questions
2014 AIPMT 2 questions
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When 0.1 mol $MnO_4^{2-}$ is oxidised the quantity of electricity required to completely oxidise $MnO_4^{2-}$ to $MnO_4^-$ is:$MnO_4^{2-} \rightarrow MnO_4^- + e^-$ (Mn goes from +6 to +7)
0.1 mol needs 0.1 mol of electrons = 0.1 F
Charge = 0.1 × 96500 = 9650 C -
The weight of silver (at. wt. = 108) displaced by a quantity of electricity which displaces 5600 mL of $O_2$ at STP will be:Faraday's second law: $\frac{w_{Ag}}{E_{Ag}} = \frac{w_{O_2}}{E_{O_2}}$
$\frac{w_{Ag}}{108} = \frac{\frac{5600}{22400}\times32}{8}$
$w_{Ag} = 108$ g