Question

If 0.0013 moles of electrons are transferred during the oxidation of copper at 0.159 amps over 701 seconds, calculate the experimental value of Faraday’s constant. (Remember that 1 A = 1 C/s which gives 1C = 1 A∙s, and F = C/mol e-).

Answer 1

current , c = 0,159 amps

= 0.159 * (C/s) [since 1A = 1 C/s]

Time, t= 701 s

charge transferred , q= c X t

= [0.159 * C/s ] X 701 s

= 111.46 C

= 111 C

number of moles of electrons transferred in the cell = 0.0013 mol.e^-

Faraday constant = 111 C / 0.0013 mol.e^-

Faraday constant = 85384 C/mol e^-

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