Chemistry, 23.08.2019 03:00 angelinaranee15

23 which term identifies the half-reaction that occurs at the anode of an operating electrochemical cell?
(1) oxidation (3) neutralization
(2) reduction (4) transmutation

Answers: 2

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Answer from: bicaplinger7766

The correct answer is Option 1.

Explanation:

Electrochemical cell is defined as the device that is used for the conversion of the chemical energy produced in a redox reaction into the electrical energy. In this cell, oxidation reaction occurs at anode and reduction reaction occurs at cathode.

For the given options:

Option 1: Oxidation reaction

Oxidation reaction is defined as the chemical reaction where a substance looses its electrons. The oxidation state of this substance gets increased and the substance gets oxidized.

$X\rightarrow X^{n+}+ne^-$

Option 2: Reduction reaction

Reduction reaction is defined as the chemical reaction where a substance gains electrons. The oxidation state of this substance gets reduced and the substance gets reduced.

$X^{n+}+ne^-\rightarrow X$

Option 3: Neutralization reaction

Neutralization reaction is defined as the chemical reaction where an acid reacts with a base to produce a salt and water molecule.

$HX+BOH\rightarrow BX+H_2O$

Option 4: Transmutation

This is defined as the process in which a nucleus is converted to another nucleus.

From the above information, at anode, oxidation reaction occurs. Thus, the correct answer is option 1.

Answer from: Quest

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: 1 - α α α = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

$k_{c}$ = $\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)v }$

or, $k_{c}$ = $\frac{\alpha ^{2} }{(1-\alpha) }$ (at v = 1l)

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: (1 - α)/0.5 α/0.5 α/0.5 = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

$\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)0.5 }$

$\frac{\alpha ^{2} }{(1-\alpha) }$ = $\frac{\alpha ^{2} }{(1-\alpha) 0.5}$

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

Answer from: Quest

the answer is minerals.

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23 which term identifies the half-reaction that occurs at the anode of an operating electrochemical...

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