Question

Number of coulombs required to deposit $90g$ of $Al$ when the electrode reaction $A{l}^{3+}+3e\to Al$ is

• A. $9.65\times {10}^6$
• B. $8.685\times {10}^4$
• C. $9.65\times {10}^4$
• D. $6.955$

Answer 1

The correct option is A $9.65\times {10}^6$

According to equation 1 mole of $Al$ i.e 27 grams of $Al$ is deposited by 3 Faraday electricity

So 90 grams of $Al$ is deposited by

$\frac{90\times 3\times 96500}{27}=9650000$
which can be written as $9.65\times {10}^6$