Question

How long will a current of \[1\] ampere take for complete deposition of copper from litre of \[1N~~~CuS{{O}_{4}}5{{H}_{2}}O\] solution?
A.\[96500\sec \]
B.\[2\times 96500\sec \]
C.\[\left[ \dfrac{96500}{2} \right]\sec \]
D.\[\left[ \dfrac{96500}{4} \right]\sec \]

Answer 1

Hint: We know that the electrochemical equivalent of a substance is the amount of substance obtained at the electrode when current of one ampere is passed through the electrolyte for one second also the question can be solved by substituting the data given in the question, to the first law of electrolysis given by Faraday.

Complete answer: In order to answer our question, we need to know about Faraday's Law of Electrolysis. Faraday's First Law of Electrolysis: This law states that the mass of a substance which is obtained at the electrode is directly proportional to the quantity of electricity passed through the electrolyte. Faraday's Second Law of Electrolysis: The law can also be stated as follows: when same quantity of electricity is passed through different electrolytes connected in series then the masses of the substances liberated at the electrodes are in the ratio of their chemical equivalent masses (atomic mass divided by number of electrons required to form the product) or the ratio of their electrochemical equivalents. Now, let us come back to our question. We have been provided with the current, and the time. So, we can substitute these into the formula for Faraday's first law and obtain the mass of copper deposited.
Here we have one litre of one normality solution. It has been given that $1$ Ampere \[=1\] faraday \[=96500\] Coulomb.
Thus, one equivalent of the \[CuS{{O}_{4}}5{{H}_{2}}O\] is present i.e. charge present \[=1\] Faraday.
Thus by charge formula \[Q=I\times t\]
By rearranging the equation we get:
\[t=\dfrac{Q}{I}=\dfrac{96500}{1}\text{ }sec\]
$\Rightarrow t=96500\sec $
Therefore, the correct answer is option A.

Note:
Remember that the change in oxidation state that takes place in the given reaction as any mistake in doing so could result in the calculation of an incorrect amount of charge which is required for the given chemical reaction. Also, be very careful of the differences between the concepts of molar and equivalent masses of compounds.