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MCQEasyJEE 2024Faraday's Laws of Electrolysis

JEE Chemistry 2024 Question with Solution

One Faraday of electricity liberates x×10−1x \times 10^{-1} gram atom of copper from copper sulphate, xx is:

  • A

    44

  • B

    55

  • C

    66

  • D

    77

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: One Faraday of electricity is passed through copper sulphate solution.

Find: The value of xx in x×10−1x \times 10^{-1} gram atom of copper liberated.

Copper is deposited according to:

Cu2++2e−→Cu\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}

This shows that 22 moles of electrons, or 22 Faradays, are required to deposit 11 mole of copper.

Therefore,

2 Faraday→1 mol Cu2 \, \text{Faraday} \rightarrow 1 \, \text{mol Cu} 1 Faraday→0.5 mol Cu1 \, \text{Faraday} \rightarrow 0.5 \, \text{mol Cu}

Since 11 gram atom of copper corresponds to 11 mole of copper, 11 Faraday liberates 0.50.5 gram atom of copper.

Now compare with the given form:

x×10−1=0.5x \times 10^{-1} = 0.5

So,

x=0.5×10=5x = 0.5 \times 10 = 5

Therefore, the correct option is B.

Using valency directly

Given: Copper ion is Cu2+\text{Cu}^{2+}.

Find: The value of xx.

For a metal ion of valency 22, one Faraday deposits:

12\frac{1}{2}

gram atom of the metal.

Thus copper deposited by 11 Faraday is:

0.5=5×10−10.5 = 5 \times 10^{-1}

Hence,

x=5x = 5

Therefore, the correct option is B.

Common mistakes

  • Assuming 11 Faraday always deposits 11 mole of metal is wrong because the deposited amount depends on the ion valency. For Cu2+\text{Cu}^{2+}, 22 electrons are needed per copper atom, so use 22 Faradays for 11 mole of copper.

  • Confusing gram atom with gram is incorrect. The question asks for gram atom, which is the number of moles of copper, not its mass in grams. Convert the deposited amount to moles before comparing with x×10−1x \times 10^{-1}.

  • Using molar mass unnecessarily without completing the conversion can lead to error. Even if 31.75 g31.75 \, \text{g} is found, it must be divided by 63.5 g mol−163.5 \, \text{g mol}^{-1} to get 0.50.5 gram atom.

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