Question

One Faraday of electricity liberates \(x×10^{−1 }\)gram atom of copper from copper sulphate, x is ______.

Answer 1

Correct Answer: 5

To determine the value of \( x \), we start by understanding the concept of Faraday's laws of electrolysis. One Faraday corresponds to the charge of one mole of electrons, which is approximately 96485 Coulombs. Copper has a valency of 2, meaning that two moles of electrons are needed to discharge one mole of copper atoms at the cathode.

Let's calculate the gram atom of copper liberated by one Faraday. The molar mass of copper is approximately 63.5 grams per mole. For copper sulfate, the reaction can be represented as:

\[ \text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu} \] 

Given:

• 1 mole of Cu requires 2 moles of electrons (2 Faradays)
• Molar mass of Cu = 63.5 g/mol

Thus, 1 Faraday will liberate:

\(\frac{63.5}{2} = 31.75\) g of Cu

Converting grams to gram-atoms (since 1 gram-atom corresponds to 1 mole):

\(\frac{31.75}{63.5} = 0.5\) gram atom of Cu

The expression is given as \( x \times 10^{-1} \) gram atom, meaning:

\( x \times 10^{-1} = 0.5 \)

Solving for \( x \):

\( x = 0.5 \times 10 = 5 \)

This calculated value (\( x = 5 \)) falls within the expected range \([5,5]\), thereby confirming the correctness of the result.

Answer 2

\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \] 
$2 \, \text{Faraday} \rightarrow 1 \, \text{mol Cu}$ 
$1 \, \text{Faraday} \rightarrow 0.5 \, \text{mol Cu deposit}$
$0.5 \, \text{mol} = 0.5 \, \text{g atom} = 5 \times 10^{-1}$ 
\[ x = 5 \]