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4 months ago

Redox Reactions Class 11th

8.35. Assertion: ClO₄^– does not show disproportionation reaction.

Reason: In ClO₄^–, chlorine is in minimum oxidation state of +1.

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Vishal Baghel

Contributor-Level 10

(c) In ClO₄^–, chlorine is in maximum oxidation state of +7. So, it does not show the disproportionation reaction.

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Redox Reactions Class 11th

8.34. Assertion: While H₂O₂ can only act as oxidising agent in their reactions.
Reason: In H₂O₂ oxidation number of O is -1 and can vary from 0 to -2

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Vishal Baghel

Contributor-Level 10

(d) In H₂O₂ oxidation number of O = -1 and can vary from 0 to -2 (+2 is possible in OF₂). The oxidation number can decrease or increase, because of this H₂O₂ can act both asoxidising and reducing agent.

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Redox Reactions Class 11th

8.33. Assertion: The highest value of oxidation number exhibited by an atom of an element generally increases across the period in the periodic table.

Reason: The oxidation number denotes theoxidation state of an element in a compound.

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Vishal Baghel

Contributor-Level 10

(b) Both are correct statements but reason is not correct for the assertion. The highest oxidation number of a representative element is the group number for the first two groups and the group number minus 10 (following the long form of periodic table) for the other groups.

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Redox Reactions Class 11th

Assertion and Reason:

Directions:

(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

(d) Assertion is wrong statement but reason is correct statement.

8.31. Assertion: Standard hydrogen electrode is called as reversible electrode.

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Vishal Baghel

Contributor-Level 10

Reason: It can react both as an anode as well a cathode in an electrochemical cell.
Answer: (a) A standard hydrogen electrode is called a reversible electrode because it can react both as anode as well as cathode in an electrochemical cell.

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Redox Reactions Class 11th

8.30. Depict the galvanic cell in which the reaction, Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s)
takes place. Further show:
(i) Which of the electrode is negatively charged.
(ii) The carriers of current in the cell and
(iii) Individual reaction at each electrode. (Advance)

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Vishal Baghel

Contributor-Level 10

The given redox reaction can be depicted as

Zn (s) + 2Ag⁺ (aq) → Zn²⁺ (aq) + 2Ag (s)
Since Zn gets oxidised to Zn²⁺ ions, and Ag⁺ gets reduced to Ag metal, therefore,

Oxidation occurs at the zinc electrode and reduction occurs at the silver electrode. Thus, galvanic cell corresponding to the above redox reaction may be depicted as:

Zn|Zn²⁺ (aq) | Ag⁺ (aq) | Ag

(i) Zinc electrode is negatively charged because oxidation occurs at the zinc electrode (i.e. electrons accumlulate on the zinc electrode)

(ii) The ions carry current. The electrons flow from Zn to Ag electrode while the current flows from Ag to Zn electrode.

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Redox Reactions Class 11th

8.29. Given the standard electrode potentials,
K⁺/K = -2.93 V, Ag⁺/Ag = 0.80 V, Hg²⁺/Hg = 0.79 V, Mg²⁺/Mg = -2.37 V,
Cr³⁺/Cr = -0.74 V. Arrange these metals in increasing order of their reducing power.

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Vishal Baghel

Contributor-Level 10

Lower the electrode potential, better is the reducing agent. Since the electrode potentials increase in the order: K⁺/K (-2.93 V), Mg²⁺/Mg (-2.37 V), Cr³⁺/Cr (-0.74 V), Hg²⁺/Hg (0.79 V), Ag⁺/Ag (0.80 V), therefore, reducing power of metals decreases in the same order, i.e., K, Mg, Cr, Hg, Ag.

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4 months ago

Redox Reactions Class 11th

8.28. Arrange the following metals in the order in which they displace each other from the solution of their salts.

Al, Cu, Fe, Mg and Zn.

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Vishal Baghel

Contributor-Level 10

Based on the relative positions of these metals in the activity series, the correct order is Mg, Al, Zn, Fe, Cu.

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Redox Reactions Class 11th

8.27. Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO₃ with silver electrodes.
(ii) An aqueous solution of silver nitrate with platinum electrodes.
(iii) A dilute solution of H₂SO₄with platinum electrodes.
(iv) An aqueous solution of CuCl₂with platinum electrodes.

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Vishal Baghel

Contributor-Level 10

(i) An aqueous solution of AgNO₃? with silver electrodes.
At cathode: Silver ions have lower discharge potential than hydrogen ions. Hence, silver ions will be deposited in preference to hydrogen ions.

At anode: Silver anode will dissolve to form silver ions in the solution.
Ag → Ag⁺ + e⁻

(ii) An aqueous solution of AgNO_3? with platinum electrodes.

At cathode: Silver ions have lower discharge potential than hydrogen ions. Hence, silver ions will be deposited in preference to hydrogen ions.

At anode: Hydroxide ions having lower discharge potential will be discharged in preference to nitrate ions. Hydroxide ions wil

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Redox Reactions Class 11th

8.26. Using the standard electrode potentials given in 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 Fe³⁺(aq). (Advance)

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Vishal Baghel

Contributor-Level 10

(i) When emf is positive, the reaction is feasible.
E^o_{? cell} = E^o_{? Fe}? −E^o_{? I2}? =0.77−0.54=0.23V
Reaction is feasible.

(ii) E^o_?cell? =E^o_{? Ag}? −E^o_{? Cu}? =0.80−0.34=0.46V
Reaction is feasible.

(iii) E^o_?cell? =E^o_{? Fe}? −E^o_{? Cu}? =0.77−0.34 =0.43 V
Reaction is feasible.

(iv) E^o_?cell? =E^o_{? Fe}? −E^o_{? Ag}? =0.77−0.80=−0.03 V
Reaction is not feasible.

(v) E^o_?cell? =E^o_{? Br2?}? −E_Fe? =1.09−0.77=0.32 V
Reaction is feasible.

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Redox Reactions Class 11th

8.25. In Ostwald's process for the manufacture of nitric add, 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 10.0 g of ammonia and 20.0 g of oxygen?

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Vishal Baghel

Contributor-Level 10

The balanced equation for the reaction is:

4NH_3? (g)+5O₂? (g)→4NO (g)+6H₂? O (g)

The molar masses of ammonia and oxygen are 17 g/mol and 32 g/mol respectively.

68 g of NH₃ will react with O₂ = 160 g.

Therefore, 10 g of NH₃ will react with O₂ = 160/68 x 10 g = 23.6 g
But the amount of O₂ which is actually available is 20.0 g which is less than the amount which is needed. Therefore, O₂ is the limiting reagent and hence calculations must be based upon the amount of O₂ taken and not on the amount of NH₃ taken. From the equation,
160 g of O₂ produce NO = 120 g
Therefore, 20 g of O₂ will produce NO =120/160 x 20 = 15 g

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