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

1.34 hr

Explanation:

From the question given above, the following data were obtained:

Number of mole of electron (e) = 0.1 mole

Current (I) = 2 A

Time (t) =?

Next, we shall determine the quantity of electricity transferred. This can be obtained as follow:

1 mole of electron = 96500 C

Therefore,

0.1 mole of electron = 0.1 × 96500

0.1 mole of electron = 9650 C

Thus, 9650 C of electricity was transferred.

Next, we shall determine the time. This can be obtained as follow:

Current (I) = 2 A

Quantity of electricity (Q) = 9650 C

Time (t) =?

Q = It

9650 = 2 × t

Divide both side by 2

t = 9650 / 2

t = 4825 s

Finally, we shall convert 4825 s to hour. This can be obtained as follow:

3600 s = 1 hr

Therefore,

4825 s = 4825 s× 1 hr / 3600 s

4825 s = 1.34 hr

Thus, the time taken is 1.34 hr

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