Question

When a solution of ${CuSO_4}$ is electrolysed for 15 minutes with a current of 1 amp, then the mass of Cu deposited at the cathode will be

• A. 30 gm
• B. 0.3 gm
• C. 0.03 gm
• D. 3 gm

Answer 1

The Correct Option is B

$I = {1.0\, A}, t = {15\, min} = 15 \times 60 { = 900\, s}$
Quantity of electricity passed = $ I \times t$
$= 1.0 \times 900$ coulombs
${= 900 \,C}$
The reaction occurring at the cathode is
${Cu^{2+} + 2e^- -> Cu}$
Thus, ${2F} i.e. 2 \times 96500 \, {C\, deposit\, Cu}$
${ = 1 \, mole = 63.5 \, g}$
and ${900\, C}$ will deposit ${Cu = \frac{63.5 }{2 \times 96500} \times 900}$
$= 0.26 \, {g}$
$\approx 0.3 \, {g}$

Answer 2

The Given data contains,

Time taken is ten minutes.

Current is 1.5 amperes.

Faraday’s constant is 96485 C mol^-1

So, let us first convert the time in minutes into seconds. This can be done by multiplying the given time duration with 60.

Now, converting time in minutes into seconds as,

Time in minutes= 10 minX\(\frac{60 sec}{{1 min}}\)

Time in minutes= 600 sec

Hence, the time taken in seconds is 600 sec.

So, let us now calculate the charge. That we can obtain by multiplying the charge and time in seconds with the current in amperes.

Therefore, Charge= Current X Time

 So, now by substituting the values of current and time in the equation of charge.

We get charge as,

Charge = Current X Time

Charge= 1.5amp X 600sec

Charge= 900C

Hence, the charge is 900C.

So, based on the following reaction,\( Cu^{2+}(aq)+2e^{-}\rightarrow Cu(s)\)

Therefore, the number of electrons that have taken part in the chemical reaction is two.

So, the molar mass of copper is 63 gmol^-1.

 So, let us now calculate the mass of copper using the values of charge, molar mass, number of electrons and Faraday's constant.

That we can obtain mass of copper deposited as,

Mass of copper deposited= \(\frac{Molarmass X Charge }{ e^{-} transferred X Faraday's Constant}\)

Mass of copper deposited= \(\frac{63 gmol^{-1} X 900 C}{2X96487 C Mol^{-1} }\)

Mass of copper deposited= \(0.2938 g\)

Therefore, the mass of copper deposited is \(0.2938 g\).

 Let us assume Faraday's constant as 96500 C Mol-1 and let the molar mass of copper be 63.5 gmol^-1.

So, we can get the mass of copper deposited as,

Mass of copper deposited= \(\frac{Molarmass X Charge }{ e^{-} transferred X Faraday's Constant}\)

Mass of copper deposited= \(\frac{63.5 gmol^{-1} X 900 C}{2X96500 C Mol^{-1} }\)

Mass of copper deposited= \(0.296 g\)

Therefore, the mass of copper deposited is \(0.296 g\) ≈\(0.3gm\).

Therefore, the correct option is ‘B’.