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3 years ago

Why are the oxidation and reduction half-reactions separated in an electrochemical cell?

Chemistry

2 answers:

Mrac [35]3 years ago

7 0

Answer:

The half-cells separate the oxidation half-reaction from the reduction half-reaction and make it possible for current to flow through an external wire.

Explanation:

ELEN [110]3 years ago

3 0

Answer:

Electrochemical cells typically consist of two half-cells. The half-cells separate the oxidation half-reaction from the reduction half-reaction and make it possible for current to flow through an external wire.

Explanation:

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This is a quiz and it’s due today

Romashka [77]

The answer is the el second one since it is a physical change

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3 years ago

Of the following equilibria, only ________ will shift to the right in response to a decrease in volume. N2 (g) + 3H2 (g) 2NH3 (g

igomit [66]

Answer:

Of the following equilibria, only one will shift to the right in response to a decrease in volume.

$N_2 (g) + 3H_2 (g)\rightleftharpoons 2NH_3 (g)$

On decreasing the volume the equilibrium will shift in right direction due to less number of gaseous moles on product side.

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

Decrease the volume

If the volume of the container is decreased , the pressure will increase according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease in pressure is taking place. So, the equilibrium will shift in the direction number of gaseous moles are less.

$N_2 (g) + 3H_2 (g)\rightleftharpoons 2NH_3 (g)$

On decreasing the volume the equilibrium will shift in right direction due to less number of gaseous moles on product side.

$2 Fe_2O_3 (s) \rightleftharpoons 4 Fe (s) + 3O_2 (g)$

On decreasing the volume the equilibrium will shift in left direction due to less number of gaseous moles on reactant side.

$2 SO_3 (g) \rightleftharpoons 2 SO_2 (g) + O_2 (g)$

On decreasing the volume the equilibrium will shift in left direction due to less number of gaseous moles on reactant side.

$H_2 (g) + Cl_2 (g) \rightleftharpoons 2 HCl (g)$

On decreasing the volume the equilibrium will shift in no direction due to same number of gaseous moles on both sides.

$2HI (g) \rightleftharpoons H_2 (g) + I_2 (g)$

On decreasing the volume the equilibrium will shift in no direction due to same number of gaseous moles on both sides.

8 0

3 years ago

A _____ (released, absorbed, taken, +,-) ΔH means that energy is______ (released, absorbed, taken, +,-) from a reaction.

zalisa [80]

A __"-"___ ΔH means that energy is_released_

4 0

4 years ago

Read 2 more answers

1) Copper oxide and carbon combine to produce metallic copper. Which of these is the balanced equation for this reaction?

Goshia [24]

The answer to your question is B

5 0

3 years ago

Express the result of the following calculation in scientific notation: 0.0263 cm2÷ 88.2 cm

VikaD [51]

<u>Answer:</u> The result of the given calculation is $2.98\times 10^{-4}cm$

<u>Explanation:</u>

Scientific notation is defined as the notation in which a number is expressed in the decimal form. This means that the number is always written in the power of 10 form.

We are given an expression:

$\Rightarrow \frac{0.0263cm^2}{88.2cm}$

The operation carried out in the above expression is division operation. The units are also treated in the same way.

The solution for the above expression is:

$\Rightarrow \frac{0.0263cm^2}{88.2cm}=0.000298cm$

The value in scientific notation is $2.98\times 10^{-4}cm$

Hence, the result of the given calculation is $2.98\times 10^{-4}cm$

3 0

4 years ago