Question

Number of coulombs required to deposit 90 gram of aluminium when the electrode reaction is Al³⁺(aq) + 3e- → Al(s) is:

1️⃣ 9.65 × 10³
2️⃣ 9.65 × 10⁵
3️⃣ 9.65 × 10⁷
4️⃣ 6.95 × 10³​

Answer 1

Given:

90 gram of aluminium is deposited.

To find:

Number of Coulombs required ?

Calculation:

$\rm \: Al^{3+}+3e^{-}\rightarrow Al$

So, from the half cell reaction, we can say that 3 moles of electrons (or 3 F electricity) is needed for 1 mole of Al (27 gram) to be deposited.

So, number of moles of electrons needed for 90 gram of Al

$= \dfrac{3}{27} \times 90$

$= 10 \: mole \: or \: faraday$

So, number of Coulombs needed are:

$= 10 \times 96500 \: C$

$= 965000 \: C$

$= 9.65 \times {10}^{5} \: C$

So, option 2) is correct !

Answer 2

Answer:

9.65×10^6

Explanation: