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In electrolysis, if the duration of the passage of current is doubled, the mass liberated is
A) Doubled
B) Halved
C) Increased four times
D) Remains the same

Answer
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Hint: The question is from the chemical effects of the current section of physics. We have to apply the concepts of electrolysis to solve this problem. We need to use Faraday’s first law and current-charge relation here.

Complete step by step solution:According to Faraday's first law, the mass of the substance released or deposited on an electrode during electrolysis is directly proportional to the amount of electric charge carried through the electrolyte.
If m = mass of a substance liberated or deposited at an electrode and q = charge. Then according to Faraday's law of electrolysis,
\(m \propto q\)
We know the relation between current and charge, which is given below.
\(I = \frac{q}{t} \Rightarrow q = I \cdot t\)
Where I = current, q = charge and t= time.
Then the Faraday's first law becomes:
\(m \propto I \cdot t\)
From this relation we can see that mass and time are directly proportional to each other. When time is doubled mass will also become double.

Hence, the correct option is Option (A).

Additional Information:Electrolysis is a technique for eliminating iron oxide. By applying a tiny electrical charge to the rusted metal from a battery or battery charger to stimulate ion exchange while the device is submerged in an electrolyte solution.
The Faraday constant represents the amount of electric charge carried by one mole, or Avogadro's number of electrons. It is a crucial constant in physics, electronics, and chemistry. The measurement is given in coulombs per mole (C/mol).

Note: According to Faraday's law of electrolysis,
\(m \propto q \Rightarrow m = zq\)
Where z = electrochemical equivalent of the substance. The z is calculated by the given equation.
\(z = \frac{{Atomic{\rm{ }}mass}}{{Valency}} \times \frac{1}{{96500}}gm{C^{ - 1}}\)