Question

32.
When a steady current of 2A was passed through two electrolytic cells A
and B containing electrolytes ZnSO, and CuSO, connected in series, 2 g of
Cu were deposited at the cathode of cell B. How long did the current flow?
What mass of Zn was deposited at cathode of cell A?
[Atomic mass : Cu = 63.5 g mol-1, Zn = 65 g mol-1; 1F = 96500 C mol-1]​

Answer 1

According to Faraday's first law of electrolysis,

mass of substance deposited is directly proportional to quantity of electricity passed.

W = ZQ = E/F * Q

2 = 31.8 / 96500 * Q

Q = 2 * 96500 / 31.8

I*t = 6069.1

t= 6069.1/ I= 6069.1/2

t = 3034.5 s

According to Faraday's second law

Mass of Fe deposited/ mass of Zn deposited = eq.wt of Fe / eq wt of Zn

2 / mass of Zn deposited = 31.8 / 32 .65

Mass of Zn deposited = 1.947 g

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Answer 2

The current did flow for 3034.5 sec and 1.947 g Zn was deposited

Given:

Current - 2A

Cells A and B containing ZnSO and CuSO series connected

2g Cu deposited at cathode cell

Atomic mass Cu = 63.5 g mol^-1

Zn = 65 g mol^-1

1F = 96500 Cmol^-1

To find:

Current flow time and Zn Deposited

Solution:

The mass of the substance deposited is directly proportional to the quantity of electricity passed

$W=ZQ= \frac{E}{F}Q$

=> $Q=\frac{2*96500}{31.8}$

$= > I*t = 6069.1 \\= > t = 3034.5 sec$

Mass of Zn deposited = $\frac{eq weight of Fe}{Eq weight of Zn}=\frac{31.8}{31.65} = 1.947 g$

The current did flow for 3034.5 sec and 1.947 g Zn was deposited

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