Redox Solved Problems

Page # 6 REDOX SOLVED PROBLEMS Ex.1 Which of the following acts as both oxidant Ex.4 Oxidation and reductant – [Co(NH3)6]Cl2Br is – (A) HNO3 (A) + 6 number of cobalt (B) Zero (C) + 3 (D) + 2 (B) HNO2 (Ans. C) (C) Both HNO3 & HNO2 Sol. (D) Neither HNO3 nor HNO2 Sol. (Ans. B) Let the O.N.of Co be x O.N. of NH3 is zero O.N. of Cl is –1 O.N. of N in HNO2 is + 3 Max. O.N. of N is + 5 O.N. of Min. O.N. of N is – 3 Hence, x + 6 (0) – 1 x 2 –1 = 0 Thus O.N. of N in HNO2 can show an increase or decrease as the case may be. That is why HNO2 acts as oxidant and reductant both. Br is –1  x = + 3 so, the oxidation number of cobalt in the given complex compound is +3. O.N. of N in HNO3 is + 5, Hence it can act only Ex.5 The order of increasing O.N. of S in as an oxidant. S8, S2O8–2 , S2O3–2, S4O6–2 is given below – Ex.2 State which of the following reactions (A) S8 < S2O8–2 < S2O3–2 < S4O6–2 is neither oxidation nor reduction – (B) S2O8–2 < S2O3–2 < S4O6–2 < S8 (A) Na  NaOH (C) S2O8–2 < S8 < S4O6–2 < S2O3–2 (B) Cl2  Cl– + ClO–3 Sol. (D) S8 < S2O3–2 < S4O6–2 < S2O8–2 (C) P2O5  H4P2O7 (D) Zn + H2SO4  ZnSO4 + H2 Sol. in (Ans. C) In the reaction P2O5  H4P2O7 The O.N. of P in P2O5 is The O.Ns. of S are shown below along with the compounds S8 , S 2O 8–2, S 2O 3–2, 0 + 6 S 4 O 6 –2 +2+2.5 Hence the order of increasing O.N. of S is 2x + 5 (–2) = 0 or x = +5 The O.N. of (Ans. D) S8 < S2O3–2 < S4O6–2 < S2O8–2 P in H4P2O7 is 4 (+1) + 2 (x) +7 (–2) = 0 Ex.6 The composition of a sample of wustite is 2x = 10 or Fe0.93O1.00. What percentage of iron is present x = +5 Since there is no change in O.N. of P, hence the above reaction is neither oxidation nor reduction. Ex.3 In the reaction in the form of Fe (III) (A) 13.05 (B) 14.05 (C) 15.05 (D) 16.05 (Ans. C) C 2O 4– 2 + MnO – 4 + H +  Mn +2 + CO 2 the reductants is – Sol. (A) C2O4–2 (B) H+ (C) MnO¯4 (D) None of the above (Ans. A) Sol. In the above reaction C 2 O 4 –2 acts as a reductant because it is oxidised to CO2 as : C2O4–2  2CO2 + 2e(oxidation) C2O4–2 reduces MnO4¯ to Mn+2 ion in solution. O.N. of. Fe in wustite is = It is an intermediate val ue 200 = 2.15 93 in between Fe (II) & Fe (III) Let % of Fe (III) be a, then 2 x (100 – a) + 3 x a = 2.15 a = 15.05  % of Fe (III) = 15.05% x100 Page # 7 REDOX Ex.7 The oxid.no. of Cl in NOClO4 is – (A) + 11 (B) + 9 (C) + 7 (D) + 5 (Ans. C) Sol. The compound may be written as NO+ ClO4¯ For ClO4–, Let Ox. No.of Cl = a Hence, the oxidation no. of Cl in NOClO4 is +7. Ex.8 The two possible oxidation numbers of N atoms in NH4NO3 are respectively – (A) +3, +5 (B) +3, –5 (C) –3, +5 (D) –3, –5 There are two N atoms in NH4NO3, but one to O). Therefore evaluation should be made separately as – N is NH4+ O.N. of N in NO3¯ a + 4 x (+1) = +1 and a + 3 (–2) = –1  a = –3  a = + 5 Ex.9 The oxidation number of S in H2S2O8 is – (C) +6 (D) +4 (Ans. C) In H2S2O8, two O atoms form peroxide linkage O O  H – O – S – O – O – S – O – H  O O S in H2S2O8 is + 6 change in oxidation number of Cr is – (C) 4 (D) 6 (Ans. A) Sol. Sol. Let the oxidation no. of S is 'a' O = –2 a = 0 Hence the oxidation no. of S in Dimethyl sulphoxide is zero. Ex.12What will be the oxidation number of I in the KI3 – (A) – 1 3 (B) – 1 4 (C) + 4 (D) + 3 Sol. In KI3 1 + 3 × (a) = 0 1 3 or KI3 is KI + I2 a = –  I has two oxidation no. –1 and 0 respectively. However factually speaking oxidation number of I Average O.N. = 1  2  ( 0 ) 1  . 3 3 respectively would be– Ex.10When K2Cr2O7 is converted into K2CrO4 the (B) 3 (Ans. C) [Fe(CN)6]–4, [Fe(SCN)]+2 and [Fe(H 2O) 6]+3  a = + 6 (A) 0 (D) 3 Ex.13Oxidation number of Fe in [Fe(CN)6]–3, 2 x 1 + 2a + 6( –2) + 2 (–1) = 0 Thus the O.N. of (B) 0 in KI3 is on average of two values – 1 and 0.   (B) 2 (Ans. A) Here the two O.N. are –3 and +5 respectivley. (B) –8 (A) 1 2(+1) + a + (–2) = 0 to H) and the other has positive oxid.no. (attached i.e. hence it is neither oxidised nor reduced and remains O.N. of N atom has negative oxidation number (attached Sol. There is no change in oxidation state of Cr, O.N. of CH3 = +1 (Ans. C) (A) +8 +6 Ex.11 The oxidation number of S in (CH3)2 SO is a = +7 O.N. of +6 in the same oxidation state. a + 4 x (–2) = –1 Sol. Cr2O7–2  CrO4–2 When Cr2O7–2 is converted into CrO4–2 the change in oxidation number of Cr is zero (A) +3, +2, +3 and +3 (B) +3, +3, +3 and +3 (C) +3, +2, +2 and +2 (D) +2, +2, +2 and +2 Sol. (Ans. A) Oxidation number of Fe in– First Second Third Fourth x–6 = –3 x–6 = –4 x–1 =+2 x+ 6×0 =+3 x = +3 x = +2 x = +3 x = +3 Page # 8 REDOX Ex.14Which of the following is not a redox Ex.17A solution containing 2.68 x 10–3 mol of reaction– A+n ions requires 1.61 x 10–3 mole of MnO–4 for the oxidation of A+n to AO–3 in acidic medium. 1 1 (A) H + I  HI 2 2 2 2 What is the value of n – (A) 1 (B) PCl5  PCl3 + Cl2 (C) 2CuSO4 + 4KI Cu2I2 + 2K2SO4 + I2 (D) CaOCl2  Ca+2 + OCl¯ + Cl¯ Sol. (Ans. D) Sol. the reactants and products remain unchanged hence, it is not a redox reaction. what is the total no. of electrons transferred during the change – (B) 24 (C) 8 (D) 12 (Ans. B) Sol. 2Alº  Al2+3  6e¯ 8e + Fe3+8/3 ............. (A) 3Feº ............. (B) Multiplying Eq. (A) by 4 and Eq. (B) by 3, then on addition 8Alº  A+n + 3H2O AO3– + 6H+ + (5–n) e– ———————————————— or 8Al + 3Fe3O4  4Al 2O3 + 9Fe Therefore, it is clear that total no. of electrons transferred during change = 24 Ex.16In the redox reaction – x KMnO4 + 24H 2SO 4  5Fe 2 (SO4)3+20CO2+y MnSO4 + 3K2SO4 + 24H2O. The values of x and y are respectively – (B) 3, 6 (C) 3, 3 (D) 6, 6 (Ans. D) Sol. The balanced redox reaction given above can be Equating these two we get 5 x 1.61 x 10–3 = (5–n) 2.68 x 10–3 Ex.18 Which of the following is correctly balanced half reaction – (A) AsO3–3+ H2O  AsO4–3 + 2H+ – 2e– (B) H2O2 + 2e  O 2 + 2H+ (C) Cr2O7–2 + 14H+  2Cr+3 + 7H2O – 6e– (D) IO–3 + 6H+  I2 + 3H2O + 5e– (Ans. C) Sol. The correctly balanced half reaction is – written as : 10FeC2O4 + 6KMnO4 + 24H2SO4 5Fe2 (SO4)3 + 20CO2 + 6 MnSO4 + 3 K2SO4 + 24H2O so the value of x = 6 and y = 6  2Cr+3 + 7H2O –6e– by ion–electron method. 8Alº + 3Fe3+8/3  9Feº + 4Al2+3 (A) 6, 3 MnO4– = 5 x 1.61 x 10–3 mol. of It is a reduction half reaction in balancing the equation  9Feº ————————————————————— 10FeC2O 4 + Amount of electrons involved in the given amount Cr2O7–2 + 14H+ 4Al2+3 + 24e 24e + 3Fe3+8/3 The reaction are  n = 2 (approx.) Ex.15In the reaction Al + Fe3O4  Al2O3 + Fe – (A) 16 (D) 4 MnO–4 + 8H+  5e–  Mn+2 + 4H2O change in oxidation states of reactant and product reaction (D) the oxidation states of the atoms of (C) 3 (Ans. B) In all the above reaction except (D) there is atoms, hence they are all redox reactions. In (B) 2