During the electrolysis of a salt of a metal M, a current of 0.5 A flows from for 32 minutes 10 seconds and deposits 0.325 g of M, What is the charge of the...
Question 1 Report
During the electrolysis of a salt of a metal M, a current of 0.5 A flows from for 32 minutes 10 seconds and deposits 0.325 g of M, What is the charge of the metal ion?(M = 65, If = 96,500 C per mole of electron)
Answer Details
Ampere = (Coulomb) x (Sec) x 1930 secs = 965 coulombs (unit of current)
But quantity of electricity = current x time = (Coulomb)/(Sec) x time
If 0.5 A for 32 mins 10 secs; Quantity of electricity = (0.5 coulomb)/(Sec x 1930) secs
= 965 coulombs
Since 965 coulombs give 0.325 g M
i.e1 coulombs = (0.325)/(965) g
∴96500 coulombs = (0.30)/(965) x 96500 =32.5 g
∴(65 g)/(32.5) = 2
But quantity of electricity = current x time = (Coulomb)/(Sec) x time
If 0.5 A for 32 mins 10 secs; Quantity of electricity = (0.5 coulomb)/(Sec x 1930) secs
= 965 coulombs
Since 965 coulombs give 0.325 g M
i.e1 coulombs = (0.325)/(965) g
∴96500 coulombs = (0.30)/(965) x 96500 =32.5 g
∴(65 g)/(32.5) = 2
inference: it takes 2 Faraday of electricity to deposit 1 mole of M. Therefore, the change on M ion = 2.
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