Q.

LIST - 1 LIST - 2
A) One faraday 1) Reduction
B) Anode 2) 96500 coulomb
C) Cathode 3) 6.24 × 1018 electrons
D) 1 coulomb 4) Oxidation
  5) Z × 96,500

 

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a

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

b

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

c

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

d

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

answer is B.

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Detailed Solution

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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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