NCERT Class 11 Chemistry Chapter 8 Redox Reactions Short Question And Answers

NCERT Solutions For Class 11 Chemistry Chapter 8 Redox Reactions Short Question And Answers

Question 1. An oxidizing agent KH(IO₃)₂ in the presence of 4.0 (N) HCI gives KCl as a product. Determine the equivalent weight of KH(IO₃)₂. [K = 39, I = 127]
Answer:

According to the question,

In the reaction, the change in oxidation number of each I -atom = + 5- (+1) = 4 units.

So, for two -atoms present in 1 molecule of KH(IO₃)₂, the total change in oxidation number

= 2 × 4 = 8 units.

So, in acid medium, the equivalent weight of KH(IO₃)₂

⇒ \(\frac{\text { Molecular weight of } \mathrm{KH}\left(\mathrm{IO}_3\right)_2}{\text { Total change in oxidation number }}\)

= \(\frac{390}{8}=48.75\)

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Question 2. Find the oxidation number of carbon in methanal and methanoic acid.
Answer:

Methanal: The oxidation number of C in the HCHO molecule = 0.

Methanol acid: Suppose the oxidation number of C in methanoic acid.

2 × (+1)(for two H-atoms)  + x + 2 × (-2) (for two O-atoms)  = 0

2 × (+1)+ x + 2 × (-2) = 0

Hence, the oxidation number of C in the HCOOH molecule = +2.

Question 3. What will be the change in the oxidation number of Mn when MnO₂ is melted with solid KNO₃ & NaOH?
Answer:

Potassium manganate is produced when MnO₂ is melted in the presence of solid KNO₃ and NaOH.

⇒ \(\stackrel{+4}{\mathrm{MnO}_2}+2 \mathrm{KOH}+\mathrm{KNO}_3 \longrightarrow \mathrm{K}_2 \stackrel{+6}{\mathrm{MnO}}{ }_4+\mathrm{KNO}_2+\mathrm{H}_2 \mathrm{O}\)

So in this reaction, the oxidation number of Mn increases from +4 to +6 i.e., a change in the oxidation number of

Mn = 6 – 4

= 2 unit

Question 4. The compound AgF₂ is unstable. However, if formed, the compound acts as a very strong oxidizing agent. Why?
Answer:

AgF₂ can be prepared, although it is not a stable compound. This is because the oxidation state of Ag in AgF₂ is +2, which is not the stable oxidation state of Ag.

The stable oxidation state of Ag is +1. As a result, Ag²⁺ in AgF₂ quickly reduces to Ag+ by gaining an electron (Ag²⁺+ e→Ag⁺). This brings about the instability of AgF₂ and makes it a very strong oxidizing agent.

Question 5. Write formulas for the following compounds:

1. Mercury chloride
2. Nickel sulphate
3. Tin 4
4. Oxide
5. Thallium sulphate Iron(3)
6. Sulphate Chromium(3) oxide

Answer:

1. HgCl₂
2. NiSO₄
3. SnO₂
4. TI₂SO₄
5. Fe₂(SO₄)₃
6. Cr₂O₃.

Question 6. A clement has three oxidation numbers, +6, +7, and +4. If it exhibits a +7 oxidation number in a compound, will the compound be able to participate In a disproportionation reaction?
Answer:

The compound cannot undergo a disproportionation reaction. This is because the element in the compound exists in its highest oxidation state. The compound would have been able to undergo a disproportionation reaction if the element existed in the +6 oxidation state in the compound, as this oxidation state lies between the oxidation states +7 and +4.

Question 7. For an element to undergo a disproportionation reaction, at least how many oxidation states should the clement exhibit?
Answer:

When an element undergoes a disproportionation reaction or oxidation state, the element changes in the following way

Intermediate- Higher oxidation + Lower oxidation

Example: The reaction of Cl₂ with cold and dilute NaOH is a disproportionation reaction.

So. an element will be able to undergo disproportionation reactionist exhibits at least three oxidation states

Question 8. An element can show 0, 1, and +5 oxidation states. The oxidation numbers of the element In two compounds are -1 and +5. Is a disproportionation reaction involving these two compounds possible?
Answer:

In a comproportionation reaction, two reactants in which a particular element exists in different oxidation states, react to form a substance in which that element exists in an intermediate oxidation state.

The given oxidation states of the element in two compounds are -1 and +5. So, if these two compounds together undergo a compo-portionation reaction, they will form a substance in which the element will exist in a zero oxidation state. This oxidation state lies between -1 and +5. Hence, the two compounds together can undergo a comproportionation reaction

Question 9. Find the oxidation state of C-l and C-2 In CH₃CH₂OH.
Answer:

The oxidation number of the three H -atoms attached to C-2 = +1. Therefore, the total oxidation number of three Hatorns = +3. For the C — C bond, the oxidation number of the C-2 atom does not change.

So, the oxidation number of the C-2 atom =-3. Now, die total oxidation number of two H -atoms attached to the C-1 atom =+2. Again, the oxidation number of the —OH group attached to the C-1 atom =-l.

Hence, the total oxidation number of two H -atoms and one linked to the C-l atom =+→2 + (-1) = +1. Thus, the oxidation number of C-1 Hence, die oxidation states of C-1 and C-2 in CH₃CH, OH are -1 and -3 respectively.

Question 10. 1 mol N₂H₄ loses 10 mol of electrons with, the formation of 1 mol of a new compound Y. If the new compound contains the same number of N-atoms then what will be the oxidation number of nitrogen in the new compound? (Assume that the oxidation number of the H -atom does not change.
Answer:

The oxidation number of each N -atom in N₂H₂ = -2 As given, 1 mol N₂H₄→1 mol Y + 10 moles.

Suppose, the oxidation of a million of N in its molecule of Y – x.

Total oxidation uninbor of two N -Moms In Y molecule -Oxidation number of two N atoms In N₂H₄ molecule = 2 × (-2) + I or, x = + 3

The oxidation number of each N-atom In compound Y = +3

Question 11. Fluorine reacts with ice and results in the change: H₂O(s) + F₂(g)-+HF(g) + HOF(g); Justify that this reaction is a redox reaction.
Answer:

⇒ \(\stackrel{+1}{\mathrm{H}}_2 \stackrel{-2}{\mathrm{O}}^{-2}(s)+\stackrel{0}{\mathrm{~F}}{ }_2(g) \rightarrow \stackrel{+1}{\mathrm{H}} \stackrel{-1}{\mathrm{~F}}^{}(g)+\stackrel{+1}{\mathrm{HOF}}^{0-1}(g)\)

In the reaction, H₂O undergoes oxidation because the oxidation number,0 increases (-2 to 0 and F₂ undergoes reduction as the oxidation number of F decreases (0 to -1). Hence, it is a redox reaction.

NCERT Solutions For Class 11 Chemistry Chapter 8 Redox Reactions Warm-Up Exercise Question And Answers

Question 1. Give two examples of nitrogen-containing compounds, in one of which, the oxidation state N-atom is +1, while in the other compound, N-atoms exist in two different oxidation states.
Answer:

The Oxidation number of N in N₂O is +1. The Oxidation Numbers Of Two N-atoms in NH₄NO₃ are -3 and +5 respectively

Question 2. CO₃O₄ is an oxide of CO₃ and CO₂ If its formula is Cox(2)Co,(m)O₄, then what is the value of x and y?
Answer:

The sum of the oxidation number of the elements in a compound is equal to zero.

So, for Co₂(2)CO_y(3)O₄, 2x+3y-4 × 2 = 0

Or, 2x+ 3y = 8

The only solution for this equation is x = 1 and y = 2.

Question 3. How many electrons should A₂H₃ liberate so that In the new compound, A shows an oxidation number of \(-\frac{1}{2}\)?
Answer:

Let, A₂H₃ will liberate x electrons.

Therefore \(2 \times\left(-\frac{1}{2}\right)+3 \times(+1)=+x\)

Or, -1+3 = +x or, x = 2.

Question 4. Give an example of an oxygen-containing compound for each of the following oxidation states of oxygen: \(+1,-\frac{1}{2},-1\)
Answer:

The oxidation states of O in O₂F₂, KO₂ And H₂O₂ are \(+1,-\frac{1}{2} \text { and }-1\) respectively

Question 5. Arrange the following compounds in increasing order of the oxidation number of N. Mg₃N₂, NH₂OH, (N₂H₅)₂SO₄, [CO(NH₃)₅CI]CI₂
Answer:

⇒ \(\mathrm{Mg}_3 \mathrm{~N}_2<\left(\mathrm{N}_2 \mathrm{H}_5\right)_2 \mathrm{SO}_4<\mathrm{NH}_2 \mathrm{OH}<\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2 ;\) 

Question 6. What are the values of a and b in the given redox reaction?

⇒ \(a \mathrm{KMnO}_4+\mathrm{NH}_3 \rightarrow b \mathrm{KNO}_3+\mathrm{MnO}_2+\mathrm{KOH}+\mathrm{H}_2 \mathrm{O}\)

Answer:

a = 8,b = 3

Question 7. Write the half-reactions of the given redox reaction

⇒ \(\mathrm{UO}^{2+}+\mathrm{Cr}_2 \mathrm{O}_7^{2-}+\mathrm{H}^{+} \rightarrow \mathrm{UO}_2^{2+}+\mathrm{Cr}^{3+}+\mathrm{H}_2 \mathrm{O}\)

Answer:

⇒ \(\mathrm{UO}^{2+}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{UO}_2^{2+}+2 \mathrm{H}^{+}+2 e\)

⇒ \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 e \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}\)

Question 8. Iodine reacts with sodium sulfate, a neutral medium. Write the balanced equation of this reaction. Calculate the equivalent mass of sodium thiosulfate in this reaction. (Assume molecular mass of sodium thiosulfate =M).
Answer:

⇒ \(2 \mathrm{~S}_2 \mathrm{O}_3^{2-}+\mathrm{I}_2 \rightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}+2 \mathrm{I}^{-}\)

Equivalent mass = M.