How many electron (in mole) will discharge 2.0g of copper(ii)ion? (Cu=64)

96.485 coulombs = 1 Faraday will deposit 64/2 g = 32 g Cu ion. Therefore, it will require

96,485 x 2/32 = ? coulombs or 1/16 of Faraday = 1/16 mole of electrons.

How many electron will be needed to discharge 32of copper

To determine the number of electrons required to discharge 2.0g of copper(II) ions, we need to use the concept of Faraday's laws of electrolysis.

1. Determine the molar mass of copper(II) ions (Cu²⁺):
The molar mass of copper (Cu) is 64g/mol. Since copper(II) ions have a +2 charge, the molar mass of Cu²⁺ is the same as Cu, which is 64g/mol.

2. Calculate the number of moles of copper(II) ions:
Since 2.0g of copper(II) ions are given, we can calculate the number of moles by dividing the mass by the molar mass:
Number of moles = mass / molar mass = 2.0g / 64g/mol = 0.03125 mol

3. Apply Faraday's laws of electrolysis:
According to Faraday's first law, one mole of any ion requires one mole of electrons to discharge completely during electrolysis.

Therefore, the number of electrons required to discharge 0.03125 mol of copper(II) ions is also equal to 0.03125 mol.

Thus, 0.03125 moles of electrons will discharge 2.0g of copper(II) ions.

To determine the number of moles of copper(II) ions, we need to use the concept of molar mass and the given mass of copper(II) ions.

Given:
Mass of copper(II) ions = 2.0 g
Molar mass of copper (Cu) = 64 g/mol

First, we calculate the number of moles of copper(II) ions:

Number of moles = mass / molar mass

Number of moles = 2.0 g / 64 g/mol = 0.03125 mol

Now, we need to determine the number of electrons discharged for every mole of copper(II) ions.

In the reaction 1 mol Cu(II) + 2 e⁻ → 1 mol Cu,

We see that for every mole of copper(II) ions, 2 electrons are discharged.

So, to find the number of electrons discharged for 0.03125 mol of copper(II) ions:

Number of electrons = number of moles x number of electrons per mole

Number of electrons = 0.03125 mol x 2 e⁻/mol = 0.0625 e⁻

Therefore, 0.0625 moles of electrons will discharge 2.0 g of copper(II) ions.