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2 years ago

In an operating voltaic cell, reduction occurs

Chemistry

2 answers:

alexandr1967 [171]2 years ago

8 0

I use this to remember "an ox, red cat"

anode oxidation
cathode reduction.

so the reduction happens at the cathode

serious [3.7K]2 years ago

6 0

Answer: Option (2) is the correct answer.

Explanation:

A voltaic cell is an electrochemical cell which helps in generation of electricity by migration of ions in the process.

In a voltaic cell, both reduction and oxidation half reactions take place. In oxidation-half reaction, there is loss of electrons whereas in reduction-half reaction there is gain of electrons.

Since, it is known that an anode is negatively charged and cathode is positively charged. So, at anode oxidation takes place and at cathode reduction takes place.

Thus, we can conclude that in an operating voltaic cell, reduction occurs at the cathode.

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See attached figure for the structure.

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Consider the following system at equilibrium where H° = -87.9 kJ, and Kc = 83.3, at 500 K. PCl3(g) + Cl2(g) PCl5(g) If the VOLUM

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Answer:

The value of Kc C. remains the same.

The value of Qc C. is less than Kc.

The reaction must: A. run in the forward direction to reestablish equilibrium

The number of moles of Cl2 will B. decrease.

Explanation:

Le Chatelier's Principle states that if a system in equilibrium undergoes a change in conditions, it will move to a new position in order to counteract the effect that disturbed it and recover the state of equilibrium.

A decrease in volume causes the system to evolve in the direction in which there is less volume, that is, where the number of gaseous moles is less.

But temperature is the only variable that, in addition to influencing equilibrium, modifies the value of the constant Kc. So if the volume of the equilibrium system is suddenly decreased at constant temperature: The value of Kc remains the same.

As mentioned, if the volume of an equilibrium gas system decreases, the system moves to where there are fewer moles. In this case, being:

PCl₃(g) + Cl₂(g) ⇔ PCl₅(g)

The equilibrium in this case then shifts to the right because there is 1 mole in the term on the right, compared to the two moles on the left. So, The reaction must: A. run in the forward direction to reestablish equilibrium.

By decreasing the volume, and so that Kc remains constant, being:

$Kc=\frac{[PCl_{5} ]}{[PCl_{3}]*[Cl_{2} ]}=\frac{\frac{nPCl_{5} }{Volume} }{\frac{nPCl_{3}}{Volume}*\frac{nCl_{2} }{Volume} } =\frac{nPCl_{5}}{nPCl_{3}*nCl_{2}} *Volume$

where nPCl₅, nPCl₃ and nCl₂ are the moles in equilibrium of PCl₅, PCl₃ and Cl₂

so, the number of moles of Cl₂ should decrease.The number of moles of Cl2 will B. decrease.

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