If 2.7 moles of copper metal was deposited from a solution of Cu(NO3)2, how many electrons were involved?

To determine the number of electrons involved in the deposition of copper metal, we need to use the concept of Faraday's law of electrolysis.

Faraday's law states that the amount of substance produced or consumed during an electrolysis reaction is directly proportional to the quantity of electricity passed through the circuit.

The equation relating moles of a substance to the number of coulombs of electricity passed is:

n = Q / F

Where:
n is the number of moles
Q is the quantity of electricity in coulombs
F is the Faraday constant (96485 C/mol)

Given that 2.7 moles of copper metal were deposited, we can calculate the quantity of electricity using the equation:

Q = n * F

Substituting the values:

Q = 2.7 mol * 96485 C/mol

Calculating:

Q = 261099.5 C

Since copper(II) ions (Cu2+) need two electrons to form one mole of copper metal, we can deduce that for every mole of copper formed, two moles of electrons are involved. Therefore, the number of electrons involved can be calculated by multiplying the number of moles of copper by 2:

Number of electrons = 2 * 2.7 mol

Calculating:

Number of electrons = 5.4 mol

Therefore, 5.4 moles of electrons were involved in the deposition of 2.7 moles of copper metal from the given solution of Cu(NO3)2.

2.7moles of Cu is 2.7*Avag number of atoms..

each atom takes two electrons.

electrons= 2*2.7*avagnumber

charge=electrons*chargeonOneelectron
= 2*2.7*avagnumber*1.60217657 × 10-19

current required=charge/time

Well, I have went further than the question asked. Your teacher is too easy.