0.1 Faraday of electricity was passed through a solution of copper (ll) sulphate. The maximum weight of copper deposited on the cathode would be [Cu = 64]
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Correct Answer: Option E
Explanation:
You're correct again — the answer is:
E. 3.2 g
Here's the step-by-step explanation:
The electrochemical equivalent formula is based on Faraday’s laws of electrolysis:
\[
\text{Mass of substance deposited} = \frac{\text{Molar mass} \times \text{Quantity of electricity (Faraday)}}{n}
\]
Where:
- Molar mass of copper, Cu = 64 g/mol
- \( n = 2 \) (because Cu²⁺ + 2e⁻ → Cu, two electrons are needed to deposit one Cu atom)
- Quantity of electricity = 0.1 Faraday
So,
\[
\text{Mass} = \frac{64 \times 0.1}{2} = \frac{6.4}{2} = \boxed{3.2 \text{ g}}
\]
You're correct again — the answer is:
E. 3.2 g
Here's the step-by-step explanation:
The electrochemical equivalent formula is based on Faraday’s laws of electrolysis:
\[
\text{Mass of substance deposited} = \frac{\text{Molar mass} \times \text{Quantity of electricity (Faraday)}}{n}
\]
Where:
- Molar mass of copper, Cu = 64 g/mol
- \( n = 2 \) (because Cu²⁺ + 2e⁻ → Cu, two electrons are needed to deposit one Cu atom)
- Quantity of electricity = 0.1 Faraday
So,
\[
\text{Mass} = \frac{64 \times 0.1}{2} = \frac{6.4}{2} = \boxed{3.2 \text{ g}}
\]