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3 years ago

I NEED THIS ANSWERS ASAP! PLZ HELP ME IT WOULD BE GREATLY APPRICIATED!!!!

Chemistry

2 answers:

pochemuha3 years ago

4 0

1. A

2. B

3. A

4. A

5. A

6. C

7. A

8. A

9. B

10. A

Please mark me brainliest

KIM [24]3 years ago

3 0

1. A
2. B
3. A
4. A
5. A
6. C
7. A
8. A
9. ?
10. A

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serious [3.7K]

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3 years ago

Balance the following skeleton reaction and identify the oxidizing and reducing agents: Include the states of all reactants and

coldgirl [10]

Answer :

a. The oxidizing agent is: $BH_4^-$

b. The reducing agent is: $ClO_3^-$

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.

Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced. Thus, it will undergo reduction reaction.

The given chemical reaction is,

$BH_4^-(aq)+ClO_3^-(aq)\rightarrow H_2BO_3^-(aq)+Cl^-(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $BH_4^-\rightarrow H_2BO_3^-$

Reduction : $ClO_3^-\rightarrow Cl^-$

Now balance oxygen atom on both side.

Oxidation : $BH_4^-\rightarrow H_2BO_3^-+3H_2O$

Reduction : $ClO_3^-+3H_2O\rightarrow Cl^-$

Now balance hydrogen atom on both side.

Oxidation : $BH_4^-+4OH^-\rightarrow H_2BO_3^-+3H_2O$

Reduction : $ClO_3^-+3H_2O\rightarrow Cl^-+6OH^-$

Now balance the charge.

Oxidation : $BH_4^-+4OH^-\rightarrow H_2BO_3^-+3H_2O+4e^-$

Reduction : $ClO_3^-+3H_2O+6e^-\rightarrow Cl^-+6OH^-$

In order to balance the electrons, we multiply the oxidation reaction by 6 and reduction reaction by 4 and then added both equation, we get the balanced redox reaction.

Oxidation : $6BH_4^-+24OH^-\rightarrow 6H_2BO_3^-+18H_2O+24e^-$

Reduction : $4ClO_3^-+12H_2O+24e^-\rightarrow 4Cl^-+24OH^-$

The balanced chemical equation in acidic medium will be,

$6BH_4^-+4ClO_3^-\rightarrow 6H_2BO_3^-+6H_2O+4Cl^-$

$3BH_4^-+2ClO_3^-\rightarrow 3H_2BO_3^-+3H_2O+2Cl^-$

In the redox reaction, $ClO_3^-$ act as oxidizing agent and $BH_4^-$ act as an reducing agent.

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Potassium (K) has 19 electrons in its natural state. It loses an electron and becomes K+ and then is left with 18 electrons .

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4 years ago