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vladimir2022 [97]
3 years ago
15

PLEASE HELP!!!

Chemistry
1 answer:
Ksju [112]3 years ago
6 0

Answer : The correct option is, (B) Oxygen diffuses through the anode.

Explanation :

Hydrogen-oxygen fuel cell : As we know that the a fuel cell is a type of device that converts the chemical energy into electrical energy. In the hydrogen-oxygen fuel cell, the proton exchange membrane cell uses hydrogen and oxygen gas as a fuel.

In this cell reaction, hydrogen and oxygen gas are the reactants and water, electricity and heat are the products.

The half-cell electrode reaction are:

At anode : 2H_2(g)+4OH^-(aq)\rightarrow 4H_2O(l)+4e^-

At cathode : O_2(g)+2H_2O(l)+4e^-\rightarrow 4OH^-(aq)

That means the hydrogen undergoes reduction at anode and oxygen undergoes oxidation at cathode.

As per question, we can say that oxygen diffuses towards anode and hydrogen diffuses towards cathode.

Hence, the correct option is, (B) Oxygen diffuses through the anode.

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Read 2 more answers
1: At which temperature would a reaction withΔH = -102 kJ/mol, ΔS = -0.188 kJ/(mol×K) be spontaneous? 2: At which temperature wo
Naddik [55]

Answer:

1: At temperatures below 542.55 K

2: At temperatures above 660 K

Explanation:

Hello there!

In this case, according to the thermodynamic definition of the Gibbs free energy, it is possible to write the following expression:

\Delta G=\Delta H-T\Delta S

Whereas ΔG=0 for the spontaneous transition. In such a way, we proceed as follows:

1:

0=\Delta H-T\Delta S\\\\T=\frac{-102kJ/mol}{-0.188kJ/mol-K} \\\\T=542.55K

It means that at temperatures lower than 542.55 K the reaction will be spontaneous.

2:

0=\Delta H-T\Delta S\\\\T=\frac{132kJ/mol}{0.200kJ/mol-K} \\\\T=660K

It means that at temperatures higher than 660 K the reaction will be spontaneous.

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7 0
2 years ago
Calculate how much heat is absorbed by a sample that weighs 12 kilograms, has a specific heat of 0.231 kg/CJ, and is heated from
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Answer:

97 J

Explanation:

Step 1: Given data

  • Mass of the sample (m): 12 kg
  • Specific heat capacity (c): 0.231 J/kg.°C (this can also be expressed as 0.231 J/kg.K)
  • Initial temperature: 45 K
  • Final temperature: 80 K

Step 2: Calculate the temperature change

ΔT = 80 K - 45 K = 35 K

Step 3: Calculate the heat required (Q)

We will use the following expression.

Q = c × m × ΔT

Q = 0.231 J/kg.K × 12 kg × 35 K = 97 J

4 0
2 years ago
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