Question

a current of 4 ampere was passed for 1.5 hours through a solution of copper sulphate when 3.2 g of copper was deposited. Calculate the current efficiency.​

Answer 1

Answer:

• Current efficiency = 45 %.

Step-by-step explanation :

Given that,

• Current = 4 amperes.
• Time = 1.5 hours = 1.5 × 60 × 60 s = 5400 seconds.

Also,

• Molar mass of copper = 63.5 g/mol.

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For deposition of Cu, the reaction is :-

$\rm{Cu^{2+}+2\:e^-\longrightarrow{}Cu}$

According to the reaction, 1 mole of Cu requires charge = 2 F = 2 × 96500 C.

∴ 3.2 g Cu will require charge :-

$\implies\rm{\dfrac{2\times{}96500}{63.5}\times{}3.2\:coulombs}$

$\implies\rm{9726\:C}$

Now, quantity of electricity passed :-

⇒ Current × Time

⇒ 4 × 5400

⇒ 21600 C.

$\therefore\rm{Current\:efficiency=\dfrac{9726}{21600}\times{}100}$

$\implies\bold{45\%}$

Answer 2

Answer:

A current of 4 ampere was passed for 1.5 hours through a solution of copper sulphate when 3.2 g of copper was deposited. Calculate the current efficiency. Quantity of electricity actually passed=4×45×60C=21600C. Current efficiency=972621600×100=45%.