An electrolytic cell transferred 0.10 mol of electrons when a constant current of 2.0 A was applied. How many hours did this take?

A. 4800h
B. 80h
C. 1.3h
D. 0.75h
E. 0.012h

I will be happy to critique your thinking.

To solve this problem, we can use Faraday's law of electrolysis, which states that the amount of charge (Q) passed through an electrolytic cell is directly proportional to the number of moles of electrons (n) transferred in the reaction.

The equation that relates these quantities is:

Q = n * F

Where:
- Q is the amount of charge in coulombs (C)
- n is the number of moles of electrons
- F is Faraday's constant, which is approximately 96,485 C/mol

In this case, we are given n = 0.10 mol and the current I = 2.0 A. We want to find the time t in hours.

We can use Ohm's law, which states that the current is equal to the charge per unit time:

I = Q / t

Rearranging this equation, we get:

Q = I * t

Now we can substitute this equation into Faraday's law:

I * t = n * F

Rearranging the equation to solve for t:

t = (n * F) / I

Substituting the given values:

t = (0.10 mol * 96,485 C/mol) / 2.0 A

Calculating this, we find:

t ≈ 4,848.5 C / A

Now we need to convert this result to hours. Recall that 1 C = 1 A * 1 s. Since there are 3600 seconds in an hour, we have:

t ≈ 4,848.5 C / (2.0 A * 3600 s/h)

Simplifying further:

t ≈ 4,848.5 / (2.0 * 3600) h

t ≈ 0.6725 h

Therefore, the correct answer is C. 1.3h.

80