The amount of Cd2+ in an aqueous sample was measured with a constant current coulometric technique. If a current of 53.6 mA required 178 s to reduce all of the Cd2+ to Cd what is the number of milligrams of Cd2+ in the solution?

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The amount of Cd2+ in an aqueous sample was measured with a constant current coulometric technique.  If a current of 53.6 mA required 178 s to reduce all of the Cd2+ to Cd what is the number of milligrams of Cd2+ in the solution?

 
Expert Solution
Step 1: Introduction to the Faraday's first law of electrolysis

From Faraday's First law of electrolysis,  the amount of the substance deposited(W) at the given electrode is directly proportional to the quantity of charge(Q) passed through the electrolytic solution.

Mathematically it is given by W = ZQ

W space equals Z space i t
W space equals fraction numerator E q u i v a l e n t space w e i g h t space o f space C d over denominator F end fraction space i t
W space equals fraction numerator left parenthesis fraction numerator A t o m i c space w e i g h t space o f space C d over denominator v a l e n c y end fraction right parenthesis over denominator F end fraction space i t

where Z = electrochemical equivalent = (E/F)

E = equivalent weight of Cd

F = Faraday's constant = 96500 C

But Q = it where i = current in amperes and t =time in seconds

i = current in amperes = space 53.6 space m A space equals 53.6 cross times 10 to the power of negative 3 end exponent space A

t = time in seconds = 178 s 

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