Question

What current was at flow of the electrolysis of copper sulphate solution, if 6 grams of copper is liberated on the cathode for 50 minutes ПОМОГИТЕ СРОЧНО ПОЖАЛУЙСТА​

Answer 1

Ответ:

m=Ikt      I=m/kt      k=0,00033 кг/Кл

I=0,006/(0,00033*50*60)=0,006 А

Объяснение:

Answer 2

To calculate the current flowing during the electrolysis of copper sulfate solution, we can use Faraday's law of electrolysis. The amount of substance (in moles) liberated during electrolysis is directly proportional to the electric charge passed through the electrolyte.

The formula for Faraday's law is:

Q = n * F

Where: Q = Electric charge passed (in Coulombs) n = Number of moles of substance liberated F = Faraday's constant (approximately 96,485 Coulombs per mole of electrons)

First, we need to find the number of moles of copper (Cu) that is liberated during the electrolysis. We can use the molar mass of copper (63.55 g/mol) to do that:

Number of moles (n) = Mass of copper (m) / Molar mass of copper

n = 6 grams / 63.55 g/mol n ≈ 0.0944 moles

Next, we can calculate the electric charge (Q) using Faraday's law:

Q = n * F Q = 0.0944 moles * 96,485 C/mol Q ≈ 9,107.6 Coulombs

Now, we know the electric charge (Q) passed during the electrolysis is approximately 9,107.6 Coulombs.

Lastly, to find the current (I), we can use the formula for current:

I = Q / t

Where: I = Current (in Amperes) Q = Electric charge (in Coulombs) t = Time (in seconds)

Given that the time (t) is 50 minutes, we need to convert it to seconds:

t = 50 minutes * 60 seconds/minute t = 3,000 seconds

Now, we can calculate the current (I):

I = 9,107.6 Coulombs / 3,000 seconds I ≈ 3.036 Amperes

Therefore, the current flowing during the electrolysis of copper sulfate solution is approximately 3.036 Amperes.