First slide

Faraday's Laws

Question

LIST - 1 LIST - 2

A) One faraday 1) Reduction

B) Anode 2) 96500 coulomb

C) Cathode 3) 6.24 × 10¹⁸electrons

D) 1 coulomb 4) Oxidation

5) Z × 96,500

Moderate

A

A - 5 B - 4 C - 2 D - 3
B

A -2 B - 4 C - 1 D - 3
C

A -2 B - 4 C - 1 D - 5
D

A - 5 B - 2 C - 1 D - 3

Solution

1 Faraday is charge carried by 1 mole of electrons = 96500 Coulomb.
At anode electrons are lost always, Oxidation occurs because Oxidation involves Loss of electrons.
At cathode electrons are gained always, Reduction occurs because Reduction involves gaining of electrons.
$Number\,\,of\,\,electrons\,\,\,in\,\,1\,Coulomb = \frac{{6.02 \times {{10}^{23}}}}{{96500}} = 6.25 \times {10^{18}}$

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