Question

The amount of copper deposited on cathode, if a current of 3A is passed through aqueous CuSO4 solution for 10 minutes, will be (At. wt of Cu = 63.5)

Answer 1

Explanation:

W=z×I×T

$w = \frac{equivlalent \: weight\: }{96500} \times i \times t$

$equivalent \: weight = \frac{atomic \: weight}{n \: factor}$

Now putting values in these formulas,

$\frac{(63.5 \div 2)}{96500} \times 3 \times (10 \times 60)$

=0.59

Answer 2

Answer:0.593 grams of cu will be deposited

Explanation:

# W = Z× (i×t)

Where,

#[ i ]=current passed

#[ t ]=timein seconds for which current is passed

[ Z ]=equivlnt wght of substance/96500

#So, now finding charge.....

Q=it......... (3A) ×10mins ×(60) =1800 C

#now, finding Z.......

[Equivalent weight / 96500]........

equi weigh= molar mass of cu / valency

=63.5/2= 31.8 grams

So, now

#Z = 31.8/96500= 0.00032

#So now final conclusion.....

W = z×it =

0.00032×1800=0.59grams

#So, 0.59 grams of copper will be deposited