Write a balanced ionic equation for the reaction. 2Ag + 2H2 S04 ———-> Ag2 S04 + 2H20 + S02 8.18 Balance the following redox reactions by ion – electron method (b) (In Acidic medium) Answer: (a) Hg(II)Cl2, (b) Ni(II)SO4, (c)Sn(IV)O2 (d) T12(I)SO4, (e) Fe2(III)(S04)3, (f) Cr2(III)O3. View Answer. What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results ? from -1 to zero. Answer: (i) KMnO4 ; K(+l); Mn(+7), 0(-2) Their oxidation potentials (a) (i) It completes the internal circuit. Balance the following equations in basic medium by ion-electron method and oxidation number methods asked Oct 8, 2017 in Chemistry by jisu zahaan ( 29.7k points) redox reaction K+/K = -2.93 V, Ag+/Ag = 0.80 V, Hg2+/Hg = 0.79 V, Mg2+/Mg = -2.37 V, (c) a catalyst (d) an acid as well as an oxidant 4. Example #1: Here is the half-reaction to be considered: MnO 4 ¯ ---> Mn 2+ It is to be balanced in acidic solution. MnO2 (s) + 4HF(l) ———–> No reaction. But the amount of O2 which is actually available is 20.0 g which is less than the amount which is needed. For reactions in a basic solution, balance the charge so that both sides have the same total charge by adding an OH-ion to the side deficient in negative charge. In this reaction, you show the nitric acid in the ionic form, because it’s a strong acid. Since the oxidation potential of Ag is much higher than that of H2O, therefore, To do so, Eq. Answer: A species which loses electrons as a result of oxidation is a reducing agent. For example, HI and HBr reduce H2S04 to S02 while HCl and HF do not. Thus, Conversely, halide ions have a tendency to lose electrons and hence can act as reducing agents. (a) While H2O2 can act as oxidising as well as reducing agent in their reactions, O3 and HNO3 acts as oxidants only. P 4 (s) + O H − (a q) → P H 3 (g) + H P O 2 − (a q). Hydrogen electrode can be made. Question 19. Question 5. Dr.Bobb222 please help balance the following oxidation-reduction reactions, which occur in acidic solution, using the half-reaction method. Answer: x + 5 (0) =0 , x = 0. (i), the sign of the electrode potential as given in Table 8.1 is reversed. (d) 5. answered Feb 14 by ... Balance MnO4^- + Fe^2+ → Fe^3+ + Mn^2+ in acidic medium by ion electron method. The oxidation number of two iodine atoms forming the I2 molecule is zero while that of iodine forming the coordinate bond is -1. Since HCl is a very weak reducing agent, it can not reduce H2S04 to S02 and hence HCl is not oxidised to Cl2. Balance the following oxidation-reduction reaction, in acidic solution, by using oxidation number method. M4O2 + 4HCI ————-> M4Cl2 + Cl2 + 2H20 Thus, at cathode, either CU2+(aq) or H2O molecules are reduced. Question 18. 2. MnO4- (aq) + Br - (aq) -> MnO2 (s) + BrO3- (aq) Question 3.Which of the following is most powerful oxidizing agent in the following. But the oxidation number cannot be fractional. (ii), we have, WARNING: This is a long answer. Answer: Question 8. 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. Therefore, they are strong oxidising agents. Question 4. Topics and Subtopics in NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reactions: NCERT Solutions Class 11 ChemistryChemistry Lab ManualChemistry Sample Papers. Question 3. (i) The cost of adding an acid or the base is avoided because in the neutral medium, the base (OH- ions) are produced in the reaction itself. F2(g) + 2I–(aq) ———-> 2F–(aq) + I2(s); Cl2 (g) + 2Br–(aq) ————> 2Cl–(aq) + Br2 (Z) (a) 6CO2(g) + 12H2O(l) ————-> C6H12O6(s) + 6H2O(l) + 6O2(g) When methane is burnt in oxygen to produce CO2 and H2O the oxidation number of carbon changes by (b) Cr is negative electrode, Pt in Mn04_ acts as positive electrode. Their relative oxidising power is, however, measured in terms of their electrode potentials. Thus, the O.N. Question 2. Writing electrode potential for each half reaction from Table 8.1, we have. Question 14. CuCl2(aq) ——-> CU2+(aq) + 2Cl–(aq) Balance the following redox reactions by ion-electron method. Therefore, F2 is both reduced as well as oxidised. of O is -1. If, however, excess of O2 is used, the initially formed CO gets oxidised to CO2 in which oxidation state of C is + 4. (iii) It is an example of a redox reaction in general and a disproportionation reaction in particular. MnO4^- ----> Mn^2+ balance O by adding H2O to the other side of the arrow Answer: Let x be the O.N. at the anode, it is the Ag of the silver anode which gets oxidised and not the H2O molecules. (d) Following the procedure detailed on page 8/23, the balanced half reaction equations are: (c) 4BCl3(g) +3LiAlH4(s) ——> 2B2H6(g) + 3LiCl(s) + 3AlCl3(s) Step2. (b) Give one example of disproportionation reaction. Since the EMF for the above reaction is positive, therefore, the above reaction is feasible. Therefore, CuO is reduced to Cu but H2 is oxidised to H20. (e) Br2 (aq) and Fe3+ (aq). In other wode either H+(aq) ions or H2O molecules are reduced. Answer: (a) Cr is getting oxidised and Mn04“ is getting reduced. Although oxidation potential of H2O molecules is higher than that of Cl– ions, nevertheless, oxidation of Cl–(aq) ions occurs in preference to H2O since due to overvoltage much lower potential than -1.36 V is needed for the oxidation of H2O molecules. Cr2O72–(aq) + 14H+(aq) + 6e– ————> 2Cr3+(aq) + 7H20(l) …(ii) Balance the Following Redox Reactions by Ion-electron Method: CBSE CBSE (Science) Class 11. H2O(aq) + 2e– ——-> H2(g) + 2OH–((aq); E° = -0.83 V First Write the Given Redox Reaction. Since Br2 is a stronger oxidising agent that I2, it oxidises S of S2O32- to a higher oxidation state of +6 and hence forms SO42- ion. (a) 4. (Balance by ion electron method) (ii) Reaction of liquid hydrazine (N 2 H 4) with chlorate ion (ClO 3 –) in basic medium produces nitric oxide gas and chloride ion in gaseous state. Write balanced chemical equation for the following reactions: (i) Permanganate ion (MnO 4 –) reacts with sulphur dioxide gas in acidic medium to produce Mn 2 + and hydrogensulphate ion. The ion-electron method allows one to balance redox reactions regardless of their complexity. (b) The possible reaction between Ag+(aq) and Cu(s) is Cu(s) + 2Ag+  (aq)—> Cu2+(aq) + 2Ag(s) of​, when you move left to right in the periodic table value of electronegativity​, Lother Meyer constructed a curve to classify the elements by studying the following propertiesA. Its electrode potential is taken as 0.000 volt. Multiply 1st equation by 1 and second equation by 2. When excess of P4 is used, PCl3 is formed in which the oxidation state of P is + 3. Solution for Balance the following redox reaction in acid: MnO4 – (aq) + C2O4 2– (aq) → Mn2+ (aq) + CO2 (g) a. Balance the following redox reaction equation by the ion-electron method in a basic solution: MnO4- + I- → MnO2 + I2. (b) When cone. Cr2O72–(aq) + 3SO2(q) + 2H+(aq) ————> 2Cr3+(aq) + 3SO42-(aq) + H20(l). (b) List three measures used to prevent rusting of iron. Thus, it is a redox reaction. Here O.N. takes place. of S in H2SO5 is 2 (+1) + x + 5 (-2) = 0 or x = +8 This is impossible because the maximum O.N. (a) H3P02(aq) + 4AgNO3(aq) + 2H2O(l) ————->H3PO4(aq) + 4Ag(s) + 4HNO3(aq) Why is standard hydrogen electrode called reversible electrode? (iii) A dilute solution of H2S04with platinum electrodes. or an oxidation state of +1 compounds of I with more electronegative elements, i.e., O, F, etc.) Reducing power goes on increasing whereas oxidising power goes on dcreasing down the series. In other words, at the cathode, either Ag+(aq) ions or H2O molecules may be reduced. Here, O is removed from CuO, therefore, it is reduced to Cu while O is added to H2 to form H20, therefore, it is oxidised. We illustrate this method … Give one example. The above redox reaction can be split into the following two half reactions. (b)Balance the following equation by oxidation number method: Balance the following redox equations by half reaction method: (i) Cr2O7^2- + Fe^2+ → Cr^3+ + H2O in acidic medium ← Prev Question Next Question → 0 votes and because of the presence of d-orbitals it also exhibits +ve oxidation states of +3, +5 and +7. First, separate the equation into two half-reactions: the oxidation portion, and the reduction portion. The following reaction, written in net ionic form, records this change. The only sure-fire way to balance a redox equation is to recognize the oxidation part and the reduction part. First Write the Given Redox Reaction. and hence it acts as an oxidant only. (Use the lowest possible coefficients. (The method below is for reactions under acidic conditions. Here, a coordinate bond is formed between I2 molecule and I– ion. The structure of H2SO5 is Question Bank Solutions 9919. This is called the half-reaction method of balancing redox reactions, or the ion-electron method. This is supported by the following reactions. Important Solutions 9. The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2 and H+ ion. (d) Ne. ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2 and H+ ion. of C in cyanide ion, CN- = x – 3 = -1 or x = +2 O.N. 1. Present a balanced equation for the reaction for this redox change taking place in water. Li (Lithium). Write a balanced redox equation for the reaction. (iv) In HNO3, O.N. Conversely, both AgNO3 and CuS04 act as oxidising agent and thus oxidise H3P02to H3P04 (orthophosphoric acid) Reaction (c) suggests that [Ag(NH3)2]+ oxidises C6H5CHO (benzaldehyde) to C6H5COO– (benzoate ion) but reaction (d) indicates that Cu2+ ions cannot oxidise C6H5CHO to C6H5COO–. (e) 8. The balanced equation is "5Fe"^"2+" + "MnO"_4^"-" + "8H"^"+" → "5Fe"^"3+" + "Mn"^"2+" + "4H"_2"O". Step2. In order to do this, the half-reaction method can be used. (i) An aqueous solution of AgNO3 with silver electrodes. Oxidation number (also called oxidation state) is a measure of the degree of oxidation of an atom in a … of C in cyanate ion, CNO =x-3-2 = -lora: = +4 The four information about the reaction are: Answer: (i) In aqueous solution, AgNO3 ionises to give Ag+(aq) and NO3– (aq) ions. (c) Following the steps as in part (a), we have the oxidation half reaction as: Fe 2+ (aq) → Fe 3+ (aq) + e-And the reduction half reaction as: H 2 O 2(aq) + 2H + (aq) + 2e- → 2H 2 O (l) Multiplying the oxidation half reaction by 2 and then adding it to the reduction half reaction, we have the net balanced redox reaction as: Each of these half-reactions is balanced separately and then combined to give the balanced redox equation. Question 5. (a) Mercury (II) chloride, (b) Nickel (II) sulphate, (c) Tin (IV) oxide, (d) Thallium (b) and (d) 9. Because of the presence of seven electrons in the valence shell, I shows an oxidation state of -1 (in compounds of I with more electropositive elements such as H, Na, K, Ca, etc.) O: I-1-+ 6OH-→ I +5 O-2 3-+ 6e- R: Mn +7 O-2 4-+ e-→ Mn +6 O-2 4 2- c) Balance the oxygen atoms. H2S04 is added to an inorganic mixture containing chloride, a pungent smelling gas HCl is produced because a stronger acid displaces a weaker acid from its salt. a. MnO4- + SO2 Mn2+ + HSO4- The reaction occurs in acidic solution. Consider the reactions: (b), Question 1. Question 8. Balancing Redox Reactions: Redox equations are often so complex that fiddling with coefficients to balance chemical equations. Also suggest a technique to investigate the path of above (a) and (b) redox reactions. To fix this issue, you must add a negative charge to the equation to balance the charges. Using the standard electrode potentials given in the Table 8.1, predict if the reaction between the following is feasible: Question 6. (c) Ozone acts as an oxidising agent. (b) MnO4–(aq) + S02(g) ——-> Mn2+(aq) +H2S04–(in acidic solution) is: 0, -1, +1, +3, +5, +7. Therefore, it quickly accepts an electron to form the more stable +1 oxidation state. Answer: Standard hydrogen electrode is known as reference electrode. Click hereto get an answer to your question ️ Balance the following redox reactions by the ion - electron method in acidic medium. The O.N. 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