Question

A current of 2.6 amp was passed through $CuS{{O}_{4}}$ solution for 380 seconds. The amount of copper deposited is:
A. 0.32g
B. 0.63g
C. 6.35g
D. 3.175g

Answer 1

The question can be solved by substituting the data given in the question, to the first law of electrolysis given by Faraday.

Complete step by step 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.. If m is mass, t is time and Z is electrochemical equivalent, then $m = ZIt$. Also, $m = \dfrac{EIT}{nF}$, where F is Faraday's constant.
- 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. The dissociation of copper sulphate look like:
$CuS{{O}_{4}}\rightleftharpoons C{{u}^{2+}}+S{{O}_{4}}^{2-}$
Hence, n = 2 as there is exchange of 2 electrons in the process. Now, F = 96500, so by substituting the data, we can write:

& W=\dfrac{EIT}{F} \\\ & \Rightarrow W=\dfrac{2.6\times 380\times 31.75}{96500} \\\ & \Rightarrow W = 0.32g \\\ \end{aligned}$$ So, we obtain the mass of copper deposited as 0.32g **The correct option is option “A” .** **Note:** 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.