The oxidation number of two iodine atoms forming the I2 molecule is zero while that of iodine forming the coordinate bond is -1. Answer:Â, Question 25. Answer: The balanced equation for the reaction is: Mnemonics for Redox â OIL - RIG OIL - RIG : Oxidation is Loss â Reduction is Gain Redox Oxidation Reduction Gain in Oxygen Loss in Hydrogen Loss in Electron (s) Loss in Oxygen Gain in Hydrogen Gain in Electron (s ) 16. Why? Question 5. Further, O.N. To do so, Eq. Justify this statement giving three illustrations. In other words, at the cathode, either Ag+(aq) ions or H2O molecules may be reduced. Thus, of Cu decreases from + 2 in CuO to 0 in Cu but that of H increases from 0 in H 2 to +1 in H 2 0. Redox Reaction Part 1 (Introduction) Redox Reaction Part 2 (Oxidation Definition) (I) sulphate, (e) Iron (III) sulphate, (f) Chromium (III) oxide. Cr3+/Cr = -0.74 V. Arrange these metals in increasing order of their reducing power. Therefore, 02 is the limiting reagent and hence calculations must be based upon the amount of 02 taken and not on the amount of NH3 taken. Answer: Oxidation involves increase in O.N while reduction involves decrease in O.N. Thus, there is no fallacy about the O.N. In Na2S04 (b) The balanced half reaction equations are: 4. Answer: (a) It may be noted that for oxidation reactions, i.e., Eq. Important questions, guess papers, most expected questions and best questions from 11th Chemistry chapter 08 Redox Reactions have CBSE chapter wise important questions with solution for free download in PDF format. Question 17. (ii) Since reactions occur faster in homogeneous medium than in heterogeneous medium, therefore, alcohol helps in mixing the two reactants, i.e., KMnO4 (due to its polar nature) and toluene (because of its being an organic compound). (a) P4(s) + OH–(aq) ———> PH3(g) + H2PO2–(aq) Answer: Let the oxidation number of S in H2SO4 be x. of C4 = + 1 + 2 (+1) + x + 1 (-1) = 0 or x = -2. GR 11 CHEM M3 ACKNOWLEDGEMENT 1 Writer Evelyn Dimacale ... 2. (a) Arrange the following in order of increasing O.N of iodine: (a) Calculate the oxidation number of Similarly, I have added an âsâ in the definitions in Slides 2, 4, 9, and 11. Chapter Wise Important Questions Class 11 Chemistry. Given the standard electrode potentials, Metallic hydrides like NaH, LiH. GRADE 11 CHEMISTRY MODULE 3 . Why it is more appropriate to write these reactions as: NCERT Solutions for Chemistry â Class 11, Chapter 8: Redox Reactions âRedox Reactionsâ is the eighth chapter in the NCERT class 11 chemistry textbook. (b)Balance the following equation by oxidation number method: (b) ClO4 – does not show disproportionation reaction. Reduction half equation: Their electrode potentials are: If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Question 6. Write the O.N of all the atoms for the following well known oxidants? Answer: Question 9. In such reactions, energy is generally liberated in the form of electrical energy. As a result, O2 is liberated at the anode according to equation (iv). Calculate the oxidation number of sulphur in H2SO4 and Na2SO4. (a) -1, -1 (b) -2, -2 (c)  -1, -2 (d) +2, -2 Question 2. Redox Reactions Class 11 Notes help the students with the summary of the chapter, important points to remember and detailed explanation of important concepts and derivations for better understanding. What is meant by electrochemical series? (ii) An aqueous solution of silver nitrate with platinum electrodes. of C. Topic 11.3.2 Types of Reactions In this topic, you will identify and describe different types of chemical reactions as a ... Redox reaction Is a reaction in which one substance is reduced and the other substance is oxidized. Here O.N. of S by chemical bonding method. Further, it may be noted that whenever any half reaction equation is multiplied by any integer, its electrode potential is not multiplied by that integer. Question 8. Answer: EMF of a cell is the difference in the electrode potentials of the two electrodes in a cell when no current flows through the cell. Further show: (a) Hg2(Br03)2 (b) Br – Cl (c) KBrO4 (d) Br2 Answer: (i) C is a reducing agent while O2 is an oxidising agent. of C in cyanate ion, CNO =x-3-2 = -lora: = +4 The four information about the reaction are: (i) by 3 and Eq. Write a balanced redox equation for the reaction. (d) C6H5CHO(l) + 2Cu2+(aq) + 5OH–(aq) ———–> No change observed (b) Cs. Consider a voltaic cell constructed with the following substances: of -2 and maximum of +6. Since the electrode potential of CU2+(aq) ions is much higher than that of H2O, therefore, at the cathode, it is CU2+(aq) ions which are reduced and not H2Omolecules. H2O(aq) + 2e– ——-> H2(g) + 2OH–((aq); E° = -0.83 V Oxidation is defined as the addition of oxygen/electronegative element to a substance or rememoval of hydrogen/ electropositive element from a susbtance. Question 10. Question 3.Which of the following is most powerful oxidizing agent in the following. Thus, when an aqueous solution of AgNO3 is electrolysed using platinum electrodes, Ag+ ions from the solution get deposited on the cathode while 02 is liberated at the anode. The p-Block Elements . (iii) Na is a reducing agent while 02 is an oxidising agent. Redox reactions are balanced by ⦠which species is oxidised. (b) HCl is a weak reducing agent and can reduce H2S04to SO2and hence HCl is not oxidised to Cl2. sulphuric’acid acts as (a) While H2O2 can act as oxidising as well as reducing agent in their reactions, O3 and HNO3 acts as oxidants only. The reduction is the gain of electrons whereas oxidationis the loss of electrons. Using PowerPoint lecture notes, present redox reactions to students. of S in H2SO5. By conventional method, the O.N. Answer: Question 8. On passing electricity, CU2+(aq) ions move towards cathode and CU2+(aq) ions move towards anode. What are characteristics of electrochemical series? Count for the fallacy. Chapter 9. (b) The possible reaction between Ag+(aq) and Cu(s) is Cu(s) + 2Ag+  (aq)—> Cu2+(aq) + 2Ag(s) (b) Which are the negative and positive electrode? It is an inert gas (with high ionization enthalpy and high positive electron gain enthalpy) and hence it neither exhibits -ve nor +ve oxidation states. 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