Answer

Verified

18.3k+ views

**Hint:**To calculate the mass of copper deposited in the reaction, we have to calculate the charge using the formula $Q = It$. 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 ${\rm{CuS}}{{\rm{O}}_{\rm{4}}}$ 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 $\dfrac{{60\;{\mathop{\rm s}\nolimits} }}{{1\;\min }}$

Now, multiply 10 minutes with the conversion factor.

$\begin{array}{c}t = 10\;\min \times \dfrac{{60\;{\mathop{\rm s}\nolimits} }}{{1\;\min }}\\ = 600\;{\rm{s}}\end{array}$

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

$Q = It$ …… (1)

Where Q is the charge, I is current and t is time. We have to substitute $I = 1.5\,{\rm{A}}$ and $t = 600\;{\rm{s}}$ in the equation (1).

$\begin{array}{c}Q = It\\ = 1.5\;{\rm{A}} \times {\rm{600}}\;{\rm{s}}\\{\rm{ = 900}}\,{\rm{C}}\end{array}$

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

Now we write the cathodic reaction of copper.

${\rm{C}}{{\rm{u}}^{2 + }} + 2{e^ - } \to {\rm{Cu}}$

Now, we apply Faraday’s law of electrolysis. From the above reaction, we see that $2 \times 96500\;{\rm{C}}$ 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.

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

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 $w \propto Q$, where w is mass deposited and Q is charge.

Recently Updated Pages

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

The path difference between two waves for constructive class 11 physics JEE_MAIN

What is the difference between solvation and hydra class 11 chemistry JEE_Main

IfFxdfrac1x2intlimits4xleft 4t22Ft rightdt then F4-class-12-maths-JEE_Main

Sodium chloride is purified by passing hydrogen chloride class 11 chemistry JEE_Main

Consider the following oxyanions PO43P2O62SO42MnO4CrO4S2O52S2O72 class 11 chemistry JEE_Main

Other Pages

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

Why does capacitor block DC and allow AC class 12 physics JEE_Main

Formal charge on nitrogen and oxygen in NO3 ion are class 11 chemistry JEE_Main

Assertion Work and heat are two equivalent from of class 11 chemistry JEE_Main

How much would a 70 kg man weigh on the moon What would class 11 physics JEE_Main

The value of de Broglie wavelength of an electron moving class 12 physics JEE_Main