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Page No 274: - Chapter 8 Redox Reactions class 11 ncert solutions Chemistry - SaraNextGen [2024]


Question 8.18:

Balance the following redox reactions by ion-electron method:

(a)

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5297/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_47cc3b45.gif  (aq) + I (aq) → MnO2 (s) + I2(s) (in basic medium)

(b)

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5297/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_47cc3b45.gif  (aq) + SO2 (g) → Mn2+ (aq) + https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5297/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_1380b916.gif  (aq) (in acidic solution)

(c) H2O2 (aq) + Fe2+ (aq) → Fe3+ (aq) + H2O (l) (in acidic solution)

(d)https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5297/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m1d462c49.gif  + SO2(g) → Cr3+ (aq) + https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5297/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_18da124d.gif (aq) (in acidic solution)

Answer:

(a) Step 1: The two half reactions involved in the given reaction are:

Oxidation half reaction: https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5297/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m7ba5b569.gif

Reduction half reaction: https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5297/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_3ad2c8c5.gif

Step 2:

Balancing I in the oxidation half reaction, we have:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5297/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_435e5c81.gif

Now, to balance the charge, we add 2 e to the RHS of the reaction.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5297/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_mb2729a1.gif

Step 3:

In the reduction half reaction, the oxidation state of Mn has reduced from +7 to +4. Thus, 3 electrons are added to the LHS of the reaction.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5297/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_56b3e6d1.gif

Now, to balance the charge, we add 4 OH ions to the RHS of the reaction as the reaction is taking place in a basic medium.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5297/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m5c532376.gif

Step 4:

In this equation, there are 6 O atoms on the RHS and 4 O atoms on the LHS. Therefore, two water molecules are added to the LHS.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5297/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_516884a8.gif

Step 5:

Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2, we have:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5297/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_3429629b.gif

Step 6:

Adding the two half reactions, we have the net balanced redox reaction as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5297/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m6ff2b4a7.gif

(b)Following the steps as in part (a), we have the oxidation half reaction as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5297/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_44f62a80.gif

And the reduction half reaction as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5297/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_6ed63141.gif

Multiplying the oxidation half reaction by 5 and the reduction half reaction by 2, and then by adding them, we have the net balanced redox reaction as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5297/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_6668ace7.gif

(c) Following the steps as in part (a), we have the oxidation half reaction as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5297/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m19c103a2.gif

And the reduction half reaction as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5297/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m71806ff5.gif

Multiplying the oxidation half reaction by 2 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5297/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_32a1ffd5.gif

(d) Following the steps as in part (a), we have the oxidation half reaction as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5297/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_56ff3360.gif

And the reduction half reaction as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5297/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_10ca4897.gif

Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5297/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m2a8af4b1.gif

Question 8.19:

Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m18d6778b.gif

Answer:

(a)

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m780fa81e.png

The O.N. (oxidation number) of P decreases from 0 in P4 to –3 in PH3 and increases from 0 in P4 to + 2 inhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m2369faad.gif . Hence, P4 acts both as an oxidizing agent and a reducing agent in this reaction.

Ion–electron method:

The oxidation half equation is:

P4s →HPO2- (aq)

The P atom is balanced as:

P4s →4HPO2- (aq)

The O atom is balanced by adding 8 H2O molecules:

P4s +8H2O →4HPO2- (aq)

The H atom is balanced by adding 12 H+ ions:

P4s +8H2O →4HPO2- (aq) + 12H+

The charge is balanced by adding e as:

P4s +8H2O →4HPO2- (aq) + 12H+ + 8e-

…(i)

The reduction half equation is:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m450fbd0d.gif

The P atom is balanced as:

P4 (s)→4PH3(g)

The H is balanced by adding 12 Has:

P4 (s) +12H+→4PH3(g)

The charge is balanced by adding 12e as:

P4 (s) +12H++12e-→4PH3(g)

…(ii)

By multiplying equation (i) with 3 and (ii) with 2 and then adding them, the balanced chemical equation can be obtained as:

5P4 (s) +24H2O →12HPO2 –  +8PH3(g) +12H+

As, the medium is basic, add 12OH both sides as:

5P4 (s) +12H2O +12OH- →12HPO2 –  +8PH3(g)

This is the required balanced equation.

Oxidation number method:

Let, total no of P reduced = x

Total no of P oxidised = 4– x

P4 (s) +OH- →xPH3(g) + 4-xHPO2 –

… (i) Total decrease in oxidation number of P = x × 3 = 3x

Total increase in oxidation number of P = (4 – x) × 2 = 8 – 2x

3x = 8 – 2x x = 8/5 From (i),

5P4 (s) +5OH- →8PH3(g) +12HPO2 –

Since, reaction occures in basic medium, the charge is balanced by adding 7OH on LHS as:

5P4 (s) +12OH- →8PH3(g) +12HPO2 –

The O atoms are balanced by adding 12H2O as:

5P4 (s) + 12H2O + 12OH-→ +12HPO2 –  + 8PH3(g)

This is the required balanced equation.

(b)

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_655741ba.jpg

The oxidation number of N increases from – 2 in N2H4 to + 2 in NO and the oxidation number of Cl decreases from + 5 in https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m6f20f594.gif  to – 1 in Cl. Hence, in this reaction, N2H4 is the reducing agent and https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m6f20f594.gif  is the oxidizing agent.

Ion–electron method:

The oxidation half equation is:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_6f6acd11.gif

The N atoms are balanced as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_5d941337.gif

The oxidation number is balanced by adding 8 electrons as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_588b6684.gif

The charge is balanced by adding 8 OHions as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m7bec1b3d.gif

The O atoms are balanced by adding 6H2O as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_5a1c3afc.gif

The reduction half equation is:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m7ad0b8c3.gif

The oxidation number is balanced by adding 6 electrons as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m38a00ceb.gif

The charge is balanced by adding 6OH– ions as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m20408c4f.gif

The O atoms are balanced by adding 3H2O as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m26e296f3.gif

The balanced equation can be obtained by multiplying equation (i) with 3 and equation (ii) with 4 and then adding them as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m49793049.gif

Oxidation number method:

Total decrease in oxidation number of N = 2 × 4 = 8

Total increase in oxidation number of Cl = 1 × 6 = 6

On multiplying N2H4 with 3 and https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m6f20f594.gif  with 4 to balance the increase and decrease in O.N., we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_5937d158.gif

The N and Cl atoms are balanced as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m6de72303.gif

The O atoms are balanced by adding 6H2O as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m49793049.gif

This is the required balanced equation.

(c)

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_54556614.jpg

The oxidation number of Cl decreases from + 7 in Cl2O7 to + 3 in https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m59600dd7.gif and the oxidation number of O increases from – 1 in H2O2 to zero in O2. Hence, in this reaction, Cl2O7 is the oxidizing agent and H2O2 is the reducing agent.

Ion–electron method:

The oxidation half equation is:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m75b62d0a.gif

The oxidation number is balanced by adding 2 electrons as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m34ba299e.gif

The charge is balanced by adding 2OHions as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m4b5081c7.gif

The oxygen atoms are balanced by adding 2H2O as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_32e8ef6b.gif

The reduction half equation is:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m1834abdf.gif

The Cl atoms are balanced as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m67404b00.gif

The oxidation number is balanced by adding 8 electrons as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_294eb882.gif

The charge is balanced by adding 6OH as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m4fea97d3.gif

The oxygen atoms are balanced by adding 3H2O as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m2be5b33a.gif

The balanced equation can be obtained by multiplying equation (i) with 4 and adding equation (ii) to it as:

 https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_12dffe48.gif

Oxidation number method:

Total decrease in oxidation number of Cl2O= 4 × 2 = 8

Total increase in oxidation number of H2O2 = 2 × 1 = 2

By multiplying H2O2 and O2 with 4 to balance the increase and decrease in the oxidation number, we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_6504e66e.gif

The Cl atoms are balanced as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_5017583f.gif

The O atoms are balanced by adding 3H2O as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_185b2614.gif

The H atoms are balanced by adding 2OH and 2H2O as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5298/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_12dffe48.gif

This is the required balanced equation.

Question 8.20:

What sorts of informations can you draw from the following reaction ?

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5299/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m314051dc.gif

Answer:

The oxidation numbers of carbon in (CN)2, CN and CNO– are +3, +2 and +4 respectively. These are obtained as shown below:

Let the oxidation number of C be x.

(CN)2

2(x – 3) = 0

= 3

CN

x – 3 = –1

= 2

CNO–

x – 3 – 2 = –1

= 4

The oxidation number of carbon in the various species is:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5299/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m4e73606f.gif

It can be easily observed that the same compound is being reduced and oxidised simultaneously in the given equation. Reactions in which the same compound is reduced and oxidised is known as disproportionation reactions. Thus, it can be said that the alkaline decomposition of cyanogen is an example of disproportionation reaction.

Question 8.21:

The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2, and H+ ion. Write a balanced ionic equation for the reaction.

Answer:

The given reaction can be represented as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5300/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m63b8c43c.gif

The oxidation half equation is:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5300/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m47e70bb.gif

The oxidation number is balanced by adding one electron as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5300/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_5f061de7.gif

The charge is balanced by adding 4H+ ions as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5300/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m370fa8c7.gif

The O atoms and H+ ions are balanced by adding 2H2O molecules as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5300/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_2d7c1f01.gif

The reduction half equation is:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5300/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_38043d44.gif

The oxidation number is balanced by adding one electron as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5300/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_cab184b.gif

The balanced chemical equation can be obtained by adding equation (i) and (ii) as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5300/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m701249b9.gif

Question 8.22:

Consider the elements:

Cs, Ne, I and F

(a) Identify the element that exhibits only negative oxidation state.

(b) Identify the element that exhibits only postive oxidation state.

(c) Identify the element that exhibits both positive and negative oxidation states.

(d) Identify the element which exhibits neither the negative nor does the positive oxidation state.

Answer:

(a) F exhibits only negative oxidation state of –1.

(b) Cs exhibits positive oxidation state of +1.

(c) I exhibits both positive and negative oxidation states. It exhibits oxidation states of – 1, + 1, + 3, + 5, and + 7.

(d) The oxidation state of Ne is zero. It exhibits neither negative nor positive oxidation states.

Question 8.23:

Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.

Answer:

The given redox reaction can be represented as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5302/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_6bf84396.gif

The oxidation half reaction is:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5302/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m1e82d110.gif

The oxidation number is balanced by adding two electrons as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5302/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_6489cd50.gif

The charge is balanced by adding 4H+ ions as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5302/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m1bfdaa54.gif

The O atoms and H+ ions are balanced by adding 2H2O molecules as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5302/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m251ed592.gif

The reduction half reaction is:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5302/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_1b5f1a72.gif

The chlorine atoms are balanced as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5302/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m25373e5f.gif

The oxidation number is balanced by adding electrons

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5302/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_66ba44e2.gif

The balanced chemical equation can be obtained by adding equation (i) and (ii) as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5302/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_5d29c10c.gif

Question 8.24:

Refer to the periodic table given in your book and now Answer the following Questions:

(a) Select the possible non metals that can show disproportionation reaction.

(b) Select three metals that can show disproportionation reaction.

Answer:

In disproportionation reactions, one of the reacting substances always contains an element that can exist in at least three oxidation states.

(a) P, Cl, and S can show disproportionation reactions as these elements can exist in three or more oxidation states.

(b) Mn, Cu, and Ga can show disproportionation reactions as these elements can exist in three or more oxidation states.

Question 8.25:

In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen?

Answer:

The balanced chemical equation for the given reaction is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5304/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m37ceeffb.gif

Thus, 68 g of NH3 reacts with 160 g of O2.

Therefore, 10g of NH3 reacts with https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5304/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_6ef2e109.gif g of O2, or 23.53 g of O2.

But the available amount of O2 is 20 g.

Therefore, O2 is the limiting reagent (we have considered the amount of O2 to calculate the weight of nitric oxide obtained in the reaction).

Now, 160 g of O2 gives 120g of NO.

Therefore, 20 g of O2 giveshttps://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5304/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_dd7b245.gif g of N, or 15 g of NO.

Hence, a maximum of 15 g of nitric oxide can be obtained.

Question 8.26:

Using the standard electrode potentials given in the Table 8.1, predict if the reaction between the following is feasible:

(a) Fe3+(aq) and I(aq)

(b) Ag+(aq) and Cu(s)

(c) Fe3+ (aq) and Cu(s)

(d) Ag(s) and Fe3+(aq)

(e) Br2(aq) and Fe2+(aq)

Answer:

(a) The possible reaction between https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_58076270.gif is given by,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_2a847047.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m6fb58566.gif

E° for the overall reaction is positive. Thus, the reaction between https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m7449b891.gif  and https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m6da111db.gif  is feasible.

(b) The possible reaction between https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m56d4ff12.gif is given by,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m648c774.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_b1b2882.gif

E° positive for the overall reaction is positive. Hence, the reaction between https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m40b1d3b.gif  and https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m3888b4a5.gif  is feasible.

(c) The possible reaction between https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m7449b891.gif and https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m3888b4a5.gif is given by,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m135808ad.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_13f05743.gif

E° positive for the overall reaction is positive. Hence, the reaction between https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m7449b891.gif  and https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m3888b4a5.gif  is feasible.

(d) The possible reaction betweenhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m387d42b7.gif and https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m7449b891.gif is given by,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_737febbd.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m19cc2958.gif

Here, E° for the overall reaction is negative. Hence, the reaction between https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m387d42b7.gif and https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m7449b891.gif  is not feasible.

(e) The possible reaction betweenhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_79074fc8.gif and https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_35b83fcb.gif is given by,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m3b959b74.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_m1e2e5228.gif

Here, E° for the overall reaction is positive. Hence, the reaction between https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_79074fc8.gif and https://img-nm.mnimgs.com/img/study_content/curr/1/11/13/201/5305/NCERT-Solutions_08-1108_Utpal_11_Chemstry_8_30_SJT_html_35b83fcb.gif  is feasible.

Also Read : Page-No-275:-Chapter-8-Redox-Reactions-class-11-ncert-solutions-Chemistry

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