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Question: What is electrolysis? A solution of \({\rm{CuS}}{{\rm{O}}_{\rm{4}}}\) is electrolysed for 10 minutes...

What is electrolysis? A solution of CuSO4{\rm{CuS}}{{\rm{O}}_{\rm{4}}} is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?.

Explanation

Solution

To calculate the mass of copper deposited in the reaction, we have to calculate the charge using the formula Q=ItQ = It. Then, using the cathodic reaction of copper we can calculate the mass of deposited copper in cathode.

Complete step by step answer:
Electrolysis is a process which involves chemical change of electrodes on passing electricity through electrolyte.

Given a solution of copper sulphate CuSO4{\rm{CuS}}{{\rm{O}}_{\rm{4}}} is electrolysed for 10 minutes with a current of 1.5 ampere. So, time for electrolysis is 10 minutes. We have to convert time from minutes to seconds. So, the conversion is factor is 60  s1  min⁡\dfrac{{60\;{\mathop{\rm s}\nolimits} }}{{1\;\min }}

Now, multiply 10 minutes with the conversion factor.
t=10  min⁡×60  s1  min⁡ =600  s\begin{array}{c}t = 10\;\min \times \dfrac{{60\;{\mathop{\rm s}\nolimits} }}{{1\;\min }}\\\ = 600\;{\rm{s}}\end{array}

Now, we have to calculate the charge for the reaction using the value of current and time. The formula of charge is,

Q=ItQ = It …… (1)

Where Q is the charge, I is current and t is time. We have to substitute I=1.5 AI = 1.5\,{\rm{A}} and t=600  st = 600\;{\rm{s}} in the equation (1).
Q=It =1.5  A×600  s=900 C\begin{array}{c}Q = It\\\ = 1.5\;{\rm{A}} \times {\rm{600}}\;{\rm{s}}\\\\{\rm{ = 900}}\,{\rm{C}}\end{array}

So, in the reaction 900  C900\;{\rm{C}} of charge passed.

Now we write the cathodic reaction of copper.
Cu2++2e−→Cu{\rm{C}}{{\rm{u}}^{2 + }} + 2{e^ - } \to {\rm{Cu}}

Now, we apply Faraday’s law of electrolysis. From the above reaction, we see that 2×96500  C2 \times 96500\;{\rm{C}} is required to deposit 63.5 grams of copper.

So, the amount of copper deposited by 900 Coulomb of charge can be calculated by the unitary method.

Amount  of  copper deposited=63.5 g2×96500 C×900  C=0.296 g\begin{array}{c}{\rm{Amount}}\;{\rm{of}}\;{\rm{copper\, deposited}} = \dfrac{{63.5\,{\rm{g}}}}{{2 \times 96500\,{\rm{C}}}} \times 900\;{\rm{C}}\\\\{\rm{ = 0}}{\rm{.296}}\,{\rm{g}}\end{array}

So, 0.296 gram of copper is deposited at cathode when copper sulphate solution is electrolysed for 10 minutes.

Note: Faraday’s law of electrolysis states that the mass of a substance liberated or deposited in an electrode is directly proportional to the quantity of charge passed. The mathematical representation of this is w∝Qw \propto Q, where w is mass deposited and Q is charge.