Question

Q.A current of 0.5 ampere is passed for 30 minutes through a voltmeter containing copper sulphate solution.Calculate the mass of Cu deposited ata the cathode .Given that atomic mass of Cu is 63-0 amu.Please give me answer fast...❤❤❤❤​

Answer 1

Answer:

0.296 grams

Explanation:

Given:

Current (I) = 0.5 A

Time taken for current to flow through Copper Sulphate Solution (t) = 30 mins

0.5 A = 0.5 C/s

⇒ Total charge (Q) = (0.5) (30 min) (60 sec) = 900 Coulomb

∴ the number of moles of Copper (n) = Q/zF

where;

z = number of electrons in the half-cell reaction

F = Faraday constant = 96,485/mol.

Hence,

n = 900/(2*96485)

= 0.00466 mol

∴ the amount of Cu deposited on the cathode = 0.00466 * 63.546

= 0.296 grams

Answer 2

mass of Cu deposited is 0.2937 g

•According to Faraday's first law :

weight deposited at cathode or anode is directly proportional to amount of charge passed through it.

• w = Zq or w = ZIt

• w= (M/nF)It where Z = (M/nF)

• where M = atomic mass = 63

• n = no. of electrons n=2

• F = Faraday's constant = 96500

• I = current passed = 0.5 A

• t =time for which current is passed

• w = (63/2×96500)×0.5×1800

• w = (63/2×96500)×900

• w = 63×9/(2×965)

• w =567 /(1930)

• w = 0.2937 g