Question

A current of 0.5A is passed for 30min through a voltmeter containing copper sulphate solution. calculate the mass of Cu deposited at the cathode. given that atomic mass of cu is 63.0u​

Answer 1

0.293 grams are deposited at the cathode

Explanation:

Given

Current I = 0.5 A

time t = 30 min ⇒ 1800 s

mass of Cu deposited = ??

Solution

we can find charges

I = Q / t

total charge in 30 min is Q = 0.5 A x 1800 s = 900 C

Then, the number of moles of Cu deposited (n) :

n = Q/zF

where, F is the Faraday constant = 96,485/mol  

z is the no. of electrons in the half-cell reaction

(z = 2)  

n = 900/(2x96485)

n = 0.00466 mol

for grams

grams = moles X molar mass

grams = 63.0 x 0.00466

grams = 0.293 g

0.293 grams are deposited at the cathode