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suter [353]
3 years ago
13

Describe the reactions during the electrolysis of water in an electrolytic cell. Group of answer choices Oxygen and hydrogen are

both oxidized. Oxygen is oxidized and hydrogen is reduced. Oxygen is reduced and hydrogen is oxidized. Oxygen and hydrogen are both reduced. Neither oxygen or hydrogen are oxidized or reduced.
Chemistry
1 answer:
Ann [662]3 years ago
6 0

Answer:

Oxygen is oxidized and hydrogen is reduced.

Explanation:

Let's consider the redox reaction during the electrolysis of water in an electrolytic cell.

2 H₂O ⇒ 2 H₂ + O₂

The corresponding half-reactions are:

2 e⁻ + 2 H₂O ⇒ H₂ + 2 OH⁻

2 H₂O ⇒ O₂ + 4 H⁺ + 4 e⁻

Oxygen is oxidized since its oxidation number increases from -2 to 0.

Hydrogen is reduced since its oxidation number decreases from +1 to 0.

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Answer:

0.013 mole.

Explanation:

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

Time (t) = 94.6 mins

Current (I) = 224.8 mA.

Mole of electrons (e) =?

Next, we shall convert 94.6 mins to seconds. This is illustrated below:

1 min = 60 s

Therefore,

94.6 mins = 94.6 min × 60 s / 1 min

94.6 mins = 5676 s

Thus, 94.6 mins is equivalent to 5676 s.

Next, we shall convert 224.8 mA to ampere (A). This can be obtained as follow:

1 mA = 1×10¯³ A

Therefore,

224.8 mA = 224.8 mA × 1×10¯³ A / 1 mA

224.8 mA = 0.2248 A

Thus, 224.8 mA is equivalent to 0.2248 A.

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

Time (t) = 5676 s

Current (I) = 0.2248 A

Quantity of electricity (Q) =?

Q = it

Q = 0.2248 × 5676

Q = 1275.96 C

Finally, we shall determine the mole of electron by converting the quantity of electricity (i.e 1275.96 C) to number of electrons. This can be obtained as follow:

96500 C = 1 e

Therefore,

1275.96 C = 1275.96 C × 1 e / 96500 C

96500 C = 0.013 e

Thus, the mole of electron trasfered in the process is 0.013 mole.

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