Question

Calculate the time required to deposit 1.27 g of copper at cathode when a current of two ampere is passed through the solution of cuso4

Answer 1

As 63.5 g of Cu is deposited by  96500C

 .’. 1.27 g of Cu is deposited by
 = 96500/63.5 x 1.27 =1930C

 As, Q = i X t [Where, Q = charge, i = current, t = time]
=> t = Q/i
Here, i = 2A, Q = 1930 C

.-. t =1930/2 = 965 s

Answer 2

The time required to deposit 1.27 g of copper at cathode when a current of two ampere is passed is 965 sec.

Faraday's law:

“The mass of a substance deposited at any electrode is without delay proportional to the quantity of charge passed.

From faraday's law of electrolysis, we get

$\mathrm{W}=\mathrm{Zit}=\frac{\mathrm{E}}{\mathrm{F}} \text { it }$

Where, W = Deposited amount

$\mathrm{E}=$ Equivalent weight

$\mathrm{F}=$ Faraday

i =Electricity passed

$t=$ Time

Calculation for time

substitute the values in given equation

$\mathrm{W}=\mathrm{Zit}=\frac{\mathrm{E}}{\mathrm{F}} \text { it }$

$\Rightarrow 1.27=\frac{63.5}{96500} \times 2 \times \mathrm{t}$

$\Rightarrow \mathrm{t}=965 \mathrm{sec}$

Hence the time required to deposit 1.27 g of copper at cathode when a current of two ampere is passed is 965 sec.

Learn more about the faraday's law here

https://brainly.in/question/17655985?msp_poc_exp=4

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