Molten calcium chloride CaCl_2CaCl2 will dissociate into its corresponding ions according to the following equation:
CaCl_(2)->Ca^(2+)+2Cl^(-)CaCl2→Ca2++2Cl−
The reduction reaction of calcium ion Ca^(2+)Ca2+ that happens on the cathode is:
Ca^(2+)+2e^(-)->Ca(s)Ca2++2e−→Ca(s)
Therefore, for 1mol1mol of Ca^(2+)Ca2+ to be reduced, 2mol2mol of electrons will be needed (2n_(ca^(2+))=n_(e^(-))2nca2+=ne−).
The number of mole of calcium Ca^(2+)Ca2+ could be found from its mass (m=60gm=60g):
n=m/(MM)=(60cancel(g))/(40cancel(g)/(mol))=1.5 molCa^(2+)
=>n_(e^(-))=2xx1.5=3.0"mol"e^(-)
The charge q applied could be calculated from the current I=5A and the time t in seconds as follows:
q=Ixxt=n_(e^(-))xxF where F=96485 C/("mol"e^(-)) is Faraday's constant.
=>t=(n_(e^(-))xxF)/I=(3.0cancel("mol"e^(-))xx(96485 C)/(1cancel("mol"e^(-))))/(5A)=57891s
=>t=(57891cancel(s))xx(1h)/(3600cancel(s))=16h
Here is video that explains further this topic:
Electrochemistry | Electrolysis, Electrolytic Cell & Electroplating.
