Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 100 seconds is (Given : Molar mass of Cu : 63 g mol^–1 , 1 F = 96487 C)

The correct answer is 0.315 g.

To find the mass of copper deposited during electrolysis, we can use Faraday's laws of electrolysis. Let's break down the solution step-by-step.

Step 1: Calculate the total charge (Q) passed through the solution
The relationship between current (I), time (t), and electric charge (Q) is given by the formula:

$Q=I\times t$

Given:
Current, $I=9.6487\text{A}$
Time, $t=100\text{s}$

Substituting these values, we get:

$Q=9.6487\text{A}\times 100\text{s}=964.87\text{C}$

Step 2: Write the reduction reaction of copper at the cathode
In a copper sulphate solution (CuSO₄), copper exists as copper ions (Cu²⁺). During electrolysis, these ions gain electrons at the cathode to deposit as metallic copper (Cu):

${\mathrm{Cu}}^{2+}+2e^{-}\to \mathrm{Cu}$

This reaction shows that 2 moles of electrons are required to deposit 1 mole of copper metal (Cu).

Step 3: Relate moles of electrons to charge
The charge of 1 mole of electrons is equal to 1 Faraday (F), which is given as 96487 C. Therefore, the charge required to deposit 1 mole (63 g) of copper is:

$\text{Charge for 1 mole of Cu}=2\times F=2\times 96487\text{C}$

Step 4: Calculate the mass of copper deposited (m)
Using the unitary method, we can calculate the mass of copper deposited by the passage of 964.87 C of charge:

$m=\frac{\text{Molar mass of Cu}\times Q}{2\times F}$

Substitute the given values:

$m=\frac{63\times 964.87}{2\times 96487}$

Simplify the fraction:

$m=\frac{63}{2}\times \frac{964.87}{96487}$

$m=31.5\times {10}^{-2}$

$m=0.315\text{g}$

Thus, the mass of copper deposited is 0.315 g.