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Veronika [31]
3 years ago
15

During the electrolysis of aqueous KCl solution using inert electrodes, gaseous hydrogen is evolved at one electrode and gaseous

chlorine at the other electrode. The solution around the electrode at which hydrogen gas is evolved becomes basic as the electrolysis proceeds. Which of the following responses describe or are applicable to the cathode and the reaction that occurs at the cathode? Select all that apply.
a. The positive electrode
b. The negative electrode
c. 2Cl– ⇌ Cl2 + 2e–
d. Cl2 + 2e– ⇌ 2Cl–
e. 2H2O ⇌ O2 + 4H+ + 4e–
f. 2H2O + 2e– ⇌ H2 + 2OH–
g. Electrons flow from the electrode to the external circuit
h. Electrons flow into the electrode from the external circuit
i. Oxidation
j. Reduction
Chemistry
1 answer:
Dima020 [189]3 years ago
6 0

Answer:

Following is the correct description of the problem.

Explanation:

  • Electricity flows through all the cathode(from the anode via the outer wire).
  • Reduction takes place at the cathode (electron benefit).
  • The cathode electrode seems to be positive as well as the anode negative.
  • 2H₂O + 2e → H₂ + 2OH⁻ (In this, there's really a rivalry between potassium as well as hydrogen ions, but mostly because hydrogen requires potential benefits for reduction or potassium requires negative ability to reduce hydrogen to hydrogen throughout the best possible method. Potassium oxidizes stronger than its ions).
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Balance each of the following in BASE: Show work
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Answer : The balanced chemical equation in a basic solution are,

(A) 2CrO_4^{2-}+8H_2O+3Cu\rightarrow 2Cr(OH)_3+4OH^-+3Cu(OH)_2

(B) 2NO_2+2OH^-\rightarrow NO_3^-+H_2O+NO_2^-

(C) 4Zn+7OH^-+NO_3^-+6H_2O\rightarrow 4[Zn(OH)_4]^{2-}+NH_3

(D) Br_2+12OH^-+5Br_2\rightarrow 2BrO_3^-+6H_2O+10Br^-

Explanation :

<u>(A) The given chemical reaction is :</u>

CrO_4^{2-}(aq)+Cu(s)\rightarrow Cr(OH)_3(s)+Cu(OH)_2(s)

The oxidation-reduction half reaction will be :

Oxidation : Cu\rightarrow Cu^{2+}+2e^-

Reduction : CrO_4^{2-}+4H_2O+3e^-\rightarrow Cr^{3+}+8OH^-

In order to balance the electrons, we multiply the oxidation reaction by 3 and reduction reaction by 2 then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

2CrO_4^{2-}+8H_2O+3Cu\rightarrow 2Cr^{3+}+16OH^-+3Cu^{2+}

or,

2CrO_4^{2-}+8H_2O+3Cu\rightarrow 2Cr(OH)_3+4OH^-+3Cu(OH)_2

<u>(B) The given chemical reaction is :</u>

NO_2(g)\rightarrow NO_3^-(aq)+NO_2^-(aq)

The oxidation-reduction half reaction will be :

Oxidation : NO_2+2OH^-\rightarrow NO_3^-+H_2O+1e^-

Reduction : NO_2+1e^-\rightarrow NO_2^-

The electrons in both the reactions are balanced. Now added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

2NO_2+2OH^-\rightarrow NO_3^-+H_2O+NO_2^-

<u>(C) The given chemical reaction is :</u>

Zn(s)+NO_3^-(aq)\rightarrow Zn(OH)_4^{2-}(aq)+NH_3(g)

The oxidation-reduction half reaction will be :

Oxidation : Zn+4OH^-\rightarrow [Zn(OH)_4]^{2-}+2e^-

Reduction : NO_3^-+6H_2O+8e^-\rightarrow NH_3+9OH^-

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

The balanced chemical equation in a basic solution will be,

4Zn+7OH^-+NO_3^-+6H_2O\rightarrow 4[Zn(OH)_4]^{2-}+NH_3

<u>(D) The given chemical reaction is :</u>

Br_2(l)\rightarrow BrO_3^-(aq)+Br^-(aq)

The oxidation-reduction half reaction will be :

Oxidation : Br_2+12OH^-\rightarrow 2BrO_3^-+6H_2O+10e^-

Reduction : Br_2+2e^-\rightarrow 2Br^-

In order to balance the electrons, we multiply the reduction reaction by 5 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

Br_2+12OH^-+5Br_2\rightarrow 2BrO_3^-+6H_2O+10Br^-

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3 years ago
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