5. write the balanced redox reaction for the decomposition of hydrogen peroxide (h₂o₂) into water and oxygen…

5. write the balanced redox reaction for the decomposition of hydrogen peroxide (h₂o₂) into water and oxygen gas. clearly indicate the oxidation and reduction half - reactions. (5 marks)\nsection b: application and problem - solving\n6. balance the following redox reaction in acidic medium using the half - reaction method: mno₄⁻+fe²⁺→mn²⁺+fe³⁺ (5 marks)\n7. a reaction is given below: 2al + 3cu²⁺→2al³⁺+3cu\n(a) which element is oxidized? (1 mark)\n(b) which element is reduced? (1 mark)\n(c) identify the oxidizing and reducing agents. (2 marks)\n8. calculate the oxidation number of chromium in the following compounds:\n(a) cr₂o₇²⁻\n(b) k₂cro₄ (4 marks: 2 marks each)\n9. write the balanced reaction for the combustion of methane (ch₄) with oxygen to form carbon dioxide and water. clearly indicate which species is oxidized and which is reduced. (5 marks)\n10. in an electrochemical cell, the following reaction occurs: zn + cu²⁺→zn²⁺+cu\n(a) write the oxidation and reduction half - reactions. (2 marks)\n(b) which electrode serves as the anode and which as the cathode? (2 marks)\n(c) explain why electrons flow from zinc to copper. (2 marks)

5. write the balanced redox reaction for the decomposition of hydrogen peroxide (h₂o₂) into water and oxygen gas. clearly indicate the oxidation and reduction half - reactions. (5 marks)\nsection b: application and problem - solving\n6. balance the following redox reaction in acidic medium using the half - reaction method: mno₄⁻+fe²⁺→mn²⁺+fe³⁺ (5 marks)\n7. a reaction is given below: 2al + 3cu²⁺→2al³⁺+3cu\n(a) which element is oxidized? (1 mark)\n(b) which element is reduced? (1 mark)\n(c) identify the oxidizing and reducing agents. (2 marks)\n8. calculate the oxidation number of chromium in the following compounds:\n(a) cr₂o₇²⁻\n(b) k₂cro₄ (4 marks: 2 marks each)\n9. write the balanced reaction for the combustion of methane (ch₄) with oxygen to form carbon dioxide and water. clearly indicate which species is oxidized and which is reduced. (5 marks)\n10. in an electrochemical cell, the following reaction occurs: zn + cu²⁺→zn²⁺+cu\n(a) write the oxidation and reduction half - reactions. (2 marks)\n(b) which electrode serves as the anode and which as the cathode? (2 marks)\n(c) explain why electrons flow from zinc to copper. (2 marks)

Answer

5.

Explanation:

Step1: Write un - balanced reaction

$H_2O_2\rightarrow H_2O + O_2$

Step2: Oxidation half - reaction

The oxygen in $H_2O_2$ has an oxidation state of - 1. In $O_2$, it has an oxidation state of 0. $2H_2O_2\rightarrow 2H_2O+O_2 + 2e^-$ (oxidation, oxygen is oxidized)

Step3: Reduction half - reaction

The oxygen in $H_2O_2$ is also reduced. $2H_2O_2 + 2e^-\rightarrow 2H_2O$ (reduction)

Step4: Combine half - reactions

The overall balanced redox reaction is $2H_2O_2\rightarrow 2H_2O + O_2$

6.

Explanation:

Step1: Write half - reactions

Oxidation: $Fe^{2 + }\rightarrow Fe^{3+}+e^-$ Reduction: $MnO_4^-\rightarrow Mn^{2+}$

Step2: Balance oxygen in reduction half - reaction

$MnO_4^-\rightarrow Mn^{2+}+4H_2O$

Step3: Balance hydrogen in reduction half - reaction (acidic medium)

$MnO_4^- + 8H^+\rightarrow Mn^{2+}+4H_2O$

Step4: Balance electrons in reduction half - reaction

$MnO_4^- + 8H^++5e^-\rightarrow Mn^{2+}+4H_2O$

Step5: Multiply oxidation half - reaction to balance electrons

$5Fe^{2 + }\rightarrow 5Fe^{3+}+5e^-$

Step6: Combine half - reactions

$MnO_4^- + 5Fe^{2+}+8H^+\rightarrow Mn^{2+}+5Fe^{3+}+4H_2O$

7.

(a)

Explanation:

Aluminum goes from an oxidation state of 0 to + 3. So, Al is oxidized.

Answer:

Al

(b)

Explanation:

Copper goes from an oxidation state of + 2 to 0. So, Cu is reduced.

Answer:

Cu

(c)

Explanation:

The substance that is oxidized is the reducing agent (Al), and the substance that is reduced is the oxidizing agent ($Cu^{2+}$).

Answer:

Oxidizing agent: $Cu^{2+}$, Reducing agent: Al

8.

(a)

Explanation:

Let the oxidation number of Cr be $x$. In $Cr_2O_7^{2 - }$, oxygen has an oxidation number of - 2. $2x+7\times(- 2)=-2$ $2x - 14=-2$ $2x=12$ $x = + 6$

Answer:

+6

(b)

Explanation:

Let the oxidation number of Cr be $x$. In $K_2CrO_4$, potassium has an oxidation number of + 1 and oxygen has an oxidation number of - 2. $2\times(+1)+x + 4\times(-2)=0$ $2+x - 8=0$ $x=+6$

Answer:

+6

9.

Explanation:

Step1: Write un - balanced reaction

$CH_4+O_2\rightarrow CO_2+H_2O$

Step2: Balance the reaction

$CH_4 + 2O_2\rightarrow CO_2+2H_2O$ Carbon in $CH_4$ has an oxidation state of - 4 and in $CO_2$ it is + 4, so carbon is oxidized. Oxygen in $O_2$ has an oxidation state of 0 and in $CO_2$ and $H_2O$ it is - 2, so oxygen is reduced.

Answer:

$CH_4 + 2O_2\rightarrow CO_2+2H_2O$, Oxidized: C in $CH_4$, Reduced: O in $O_2$

10.

(a)

Explanation:

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

Answer:

Oxidation: $Zn\rightarrow Zn^{2+}+2e^-$, Reduction: $Cu^{2+}+2e^-\rightarrow Cu$

(b)

Explanation:

In an electrochemical cell, the oxidation occurs at the anode and reduction at the cathode. Since Zn is oxidized, Zn is the anode and since $Cu^{2+}$ is reduced, Cu is the cathode.

Answer:

Anode: Zn, Cathode: Cu

(c)

Explanation:

Zinc has a lower reduction potential than copper. Zinc is more likely to lose electrons (oxidize) and these electrons flow through the external circuit to the copper electrode where $Cu^{2+}$ ions gain the electrons (reduce).

Answer:

Zinc has a lower reduction potential and is more easily oxidized, so electrons flow from Zn to Cu.