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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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The given reaction is as follows:

2NO (g) + O₂ (g) = 2NO₂ (g), ΔH = -114 kJ

It is known that dSsurr = -dHsys / T (Temp = 355 K)

So,  dSsurr = - (-114 × 1000) / 355

dSsurr = +321.12 J/K

Hence, the value of dSsurr is +321.12 J/K

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It is known that dG = dHsys - TdSsys,

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dG<0

Also, dStotal = dSsys + dSsurr > 0

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Thus, the assumption is correct, and the given reaction is spontaneous. Hence, the final answer is Ssurr = +321 J/K reaction is spontaneous.



8 0
3 years ago
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