Question

1.02 g of zinc were released in an electrolysis bath with a current of 5 A for 10 minutes. Calculate the electrochemical zinc equivalent

Answer 1

Given :

• Mass of zinc (m) = 1.02g
• Current (i) = 5A
• For time (t) = 10min

To Find :

• Electrochemical Equivalent (Z)

Formula Used :

Faraday First Law of electrolysis

Amount of substance deposited or liberated at an electrode is directly proportional to Amount of charge passed (utilized) through the solution

$\implies{\sf m \; \propto \: Q}$

$\implies{\sf m = Zit }$

m = mass of zinc ;

Z = Electrochemical equivalent

Q = Charge = it ; i = current ; t = time

Solution :

Electrochemical Equivalent

Weight deposited by 1 Coulomb charge is called electrochemical equivalent (Z)

$\implies{\sf m = Zit}$

$\implies{\sf 1.02 = Z \times 5 \times 10 }$

$\implies{\sf Z = \dfrac{1.02 }{5 \times 10}}$

$\implies{\sf Z =\dfrac{1.02}{50}}$

$\implies{\bf Z = 0.0204\: g/C}$

Answer :

Electrochemical equivalent is 0.0204 g/C

Answer 2

Answer:

$\huge\bf{GIVEN\:}$

1. Mass of Zinc = 1.02 g
2. Current = 5A
3. Time = 10 Min

To Find:

Electrochemical Zinc Equivalent ?

In this Question,

The Formula of Faraday First Law of Electrolysis Used :

$\implies {m \propto Q}$

$\implies {m\:=\:Zit}$

Remarks:

M = Mass of Zinc

Z = Electrochemical Zinc Equivalent

Q = Charge

i = Current

t = time

$\bold\green{Solution}$

Electrochemical Equivalent

$⟼ {m\:=\:Zit}$

$⟼ {1.02\:=\:Z\:×\:5\:×\:10}$

$⟼ {Z\:= \frac{1.02}{5\:×\:10}}$

$⟼ {Z\:= \frac{1.02}{50}}$

$⟼ {Z\:=\:<strong>0.0204</strong>\:<strong>g/C</strong>}$