Electrochemistry Solved Examples

ELECTROCHEMISTRY SOLVED EXAMPLES Ex. 1 Colour of KI solution containing starch turns blue when Cl2 water is added. Explain. Sol. Chlorine placed below iodine in electrochemical series having more reduction potential and thus shows reduction whereas I– undergoes oxidation. The I2 so formed get absorbed in starch to give blue colour. 2I–  I2 + 2e– Cl2 + 2e–  2Cl– ————————— Cl2 + 2I–  I2 + 2Cl– Ex. 2 The resistance of a 0.01 N solution of an electrolyte was found to 210 ohm at 298 K using a conductivity cell with a cell constant of 0.88 cm–1. Calculate specific conductance and equivalent conductance of solution. Given for 0.01 N solution. Sol. R = 210 ohm l = 0.88 cm–1 A Specific conductance  κ= 1  × R A κ= 1 × 0.88 = 4.19 × 10–3 mho cm–1 210 λ eq = k ×1000 N λ eq = 4.19 ×10 –3 ×1000 0.01 eq.= 419 mho cm2 eq–1. Ex. 3 The conductivity of pure water in a conductivity cell with electrodes of cross-sectional area 4 cm2 placed at a distance 2 cm apart is 8 × 10–7 S cm–1. Calculate : (A) the resistance of water. (B) the current that would flow through the cell under the applied potential difference of 1 volt. Sol. Cell constant = (A) (B) l 2 1 = = cm–11 A 4 2 Also From Ohm's law, κ= 1 l × R A R= 1 l 1 1 × = × = 6.25 × 105 ohm –7 κ A 8 ×10 2 V =R I  I= 1 = 1.6 × 10–6 ampere 6.25 ×105 Ex. 4 Sol. Ex. 5 Sol. How many moles of electrons are needed for the reduction of 20 mL of 0.5M solution of KMnO4 in acid medium ? Moles of KMnO4 = M × V (L) = 0.5 × 20 × 10–3 = 10–2 [ Mn7+ + 5e–  Mn2+]  1 mol KMnO4 required = 5 mol e–  10–2 mol KMnO4 required = 5 × 10–2 mol e– An aqueous solution of NaCl is electrolysed with inert electrodes. Write the equations for the reactions taking place at cathode and anode. What happens if NaNO3(aq.) is used instead of NaCl ? For NaCl(aq.) anode : 2Cl–  Cl2 + 2e– cathode : 2H+ + 2e–  H2 For NaNO3(aq.) anode cathode Ex. 6 Sol. : 2OH–  H2O + 1 2 O2 + 2e– : 2H+ + 2e–  H2 On electrolysing CuSO4 solution in presence of Pt, the solution becomes colourless, blue colour of solution disappears. Why ? At cathode : Cu2+ + 2e–  Cu At anode : 2OH–  H2O + 1 2 O2 + 2e– Cu2+ ions are reduced and blue colour is due to Cu2+. Ex. 7 The specific conductivity of a solution containing 1.0 g of anhydrous BaCl2 in 200 cm3 of the solution has been found to be 0.0058 S cm–1. Calculate the molar and equivalent conductivity of the solution. Molecular wt. of BaCl2 = 208. Sol. Molarity of BaCl2 = Also Ex. 8 Sol. normality of BaCl2 1×1000 = 0.024 M 208 × 200 = 0.024 × 2 = 0.048 N (N = M × V. f) Now μ = k× 1000 0.0058 ×1000 = = 241.67 S cm2 mol–1 CM 0.024 Also λ = k× 1000 0.0058 ×1000 = = 120.83 S cm2 equivalent–1 CN 0.048 The resistance of a solution 'A' is 50 ohm and that of solution 'B' is 100 ohm, both solution being taken in the same conductivity cell. If equal volumes of solutions A and B are mixed, what will be the resistance of the mixture using the same cell ? (Assume that there is no increase in the degree of dissociation of A and B on mixing.) Let k1 and k2 be the specific conductance of the solution A and B respectively and the cell constant of the cell be X. For solution A : Sp. conductance = Conductance × Cell constant 1 k1 = ×X ..................... (i) 50 1 ×X For solution B : Sp. conductance, k 2 = .................. (ii) 100 When equal volumes of A and B are mixed, both the solutions get doubly diluted, hence their individual contribution k2 k1 towards the sp. conductance of the mixture will be and respectively and the sp. 2 2 1 conductance of the mixture will be (k1 + k2). 2 1 1 (k1 + k2) = × X 2 R (R is the resistance of mixture) From equation (i), (ii) and (iii) ; R = 66.67 ohm  For the mixture : ..................... (iii) ELECTROCHEMISTRY Ex. 9 Sol. The value of  for NH4Cl, NaOH and NaCl are 129.8, 248.1 and 126.4 ohm–1 cm2 mol–1 respectively. Calculate  for NH4OH solution.  NH 4 OH   NH 4 Cl   NaOH   NaCl = 129.8 + 248.1 – 126.4 NH 4 OH = 251.5 ohm–1 cm2 mol–1 Ex. 10 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 = –0.74V. Arrange these metals in their increasing order of reducing power. Sol. More is E°RP, more is the tendency to get reduced or more is the oxidising power or lesser is reducing power. Ag < Hg < Cr < Mg < K Ex. 11 Answer of following : (i) In a cell, anode is represented by –ve sign and cathode by +ve sign. What is the type of the cell? (ii) Write the reaction taking place at anode in the following cell : Pt, H2 (1atm) | HCl (1M) Cl2 , Pt. (iii) With the help of the following possible reactions, arrange Mg, Zn, Cu and Ag in decreasing order of their reduction potentials. Cu + 2Ag+  Cu2+ + 2Ag Mg + Zn2+  Mg2+ + Zn Zn + Cu2+  Zn2+ + Cu (iv) Cl2 can displace I2 from KI solution but I2 does not displace. (v) The E° (standard oxidation potentials) values of the elements A, B, C and D are + 0.76, – 0.34, + 0.25 and – 0.85 volt respectively. Arrange these in decreasing order of reactivity. Sol. (i) The cell is an electrochemical cell i.e. galvanic cell. (ii) Reaction taking place at anode is : H2  2H+ + 2e– (iii) The decreasing order of reduction potentials in : Ag > Cu > Zn > Mg (iv) I– is stronger reducing agents than Cl–. Therefore, I– can reduce Cl2 to Cl– ions whereas Cl– cannot reduct I2 to I–. Thus, only the following reaction is feasible. Cl2 + 2I– (2KI) 2KCl (2Cl–) + I2 (v) The reactivity of an elements is directly porportional to its standard oxidation potential (E°). The decreasing order of reactivity of elements is : A > C > B > D. Ex. 12 A cell is prepared by dipping a copper rod in 1 M CuSO4 solution and a nickel rod in 1M NiSO4. The standard reduction potentials of copper and nickel electrodes are + 0.34 V and –0.25 V respectively. (i) Which electrode will work as anode and which as cathode ? (ii) What will be the cell reaction ? (iii) How is the cell represented ? (iv) Calculate the EMF of the cell. Sol. (i) The nickel electrode with smaller E° value (–0.25 V) will work as anode while copper electrode with more E° value (+0.34V) will work as cathode. (ii) The cell reaction may be written as : (iii) At anode : Ni (s)  Ni2+ (aq.) + 2e– At cathode : Cu2+ (aq) + 2e– Cu (s) Cell reaction : Ni (s) + Cu2+ (aq)  Ni2+ (aq) + Cu (s) The cell may be represented as : Ni(s)/Ni2+ (aq) ||Cu2+ (aq)/Cu (s) ° ° EMF of cell = E cathode – E anode = (+0.34) – (–0.25) = 0.59 V (iv) Ex. 13 Calculate molar conductance for NH4OH, given that molar conductances for Ba(OH)2, BaCl2 and NH4Cl are 523.28, 280.0 and 129.8 ohm–1 cm2 mol–1 respectively. Sol.   μ Ba(OH)2 = λ Ba = 523.28 2+ + 2λ OH- ......... (i) μ BaCl2 = λ Ba 2+ + 2λ Cl- = 280.00 ......... (ii)  μ NH 4Cl = λ NH + + λ Cl - = 129.80 4 ........ (iii)  μ NH4OH = λ NH 4 + λ OH –  Eq.(i) Eq.(ii) will gives – 2 2 502.88   + λ OH – = λ = 251.44 ohm–1 cm2 mol–1 NH 4 OH – = 2 Eq.(iii) + λ NH+ 4 Ex. 14 The equivalent conductivities of acetic acid at 298 K at the concentrations of 0.1 M and 0.001 M are 5.20 and 49.2 S cm2 eq.–1 respectively. Calculate the degree of dissociation of acetic acid at these concentrations. Given that, (H+) and (CH3COO–) are 349.8 and 40.9 ohm–1 cm2 mol–1 respectively. Sol. Degree of dissociation is given by α = (i) λc λ  Evaluation of λ CH3COOH  λ CH3COOH = λ CH3COO – +  H  = 40.9 + 349.8 = 390.7 ohm–1 cm2 eq.–1 (ii) Evaluation of degree of dissociation At C = 0.1 M 5.20 λc = = 0.013 i.e. λ  390.7 1.3%  = 1.3% or At C = 0.001 M α= α= 49.2 λc = 0.125  = 390.7 λ i.e. 12.5% ELECTROCHEMISTRY Ex. 15 E° of some oxidants are given as : I2 + 2e– 2I– E° = + 0.54 V MnO 4  8 H  + 5e–  Mn2+ + 4H2O E° = + 1.52 V Fe3+ + e–  Fe2+ Sn + 2e  Sn 4+ – 2+ E° = + 0.77 V E° = + 0.1 V (A) Select the strongest reductant and oxidant in these. (B) Select the weakest reductant and oxidant in these. (C) Select the spontaneous reaction from the changes given below. Sol. (i) Sn4+ + 2Fe2+  Sn2+ + 2Fe3+ (ii) 2Fe2+ + I2  2Fe3+ + 2I– (iii) Sn4+ + 2I–  Sn2+ + I2 (iv) Sn2+ + I2  Sn4+ + 2I– (A) More the E°OP, more is the tendency for oxidation. Therefore, since maximum E°OP stands for : Sn2+  Sn4+ + 2e–  (B) E°OP = – 0.1V strongest reductant : Sn 2+ and weakest oxidant : Sn 4+ More or +ve is E°RP ,more is the tendency for reduction. Therefore, since maximum E°RP stands for : – 2+ MnO 4  8H  + 5e Mn + 4H2O  E°RP = + 1.52 V strongest oxidant : MnO 4 and weakest reductant : Mn2+ Note : Stronger is oxidant, weaker is its conjugate reducant and vice-versa. (C) For (i) E °Cell = E °OP Fe2+ /Fe3+ + E °RP Sn 4+ /Sn 2+ = –0.77 + 0.1 Fe2+ oxidizes and Sn4+ reduces in change.  E °Cell = – 0.67 V E °Cell is negative.  (i) For (ii) E °Cell = E °OP  (ii) Is non-spontaneous change. Fe2+ /Fe3+ + E °RP I 2 /I- = –0.77 + 0.54 = – 0.23 V Is non-spontaneous change. ° ° ° For (iii) E Cell = E OPI– /I2 + E RPSn4+ /Sn2+ – = –0.54 + 0.1 = – 0.44 V  (iii) Is non-spontaneous change. ° ° ° For (iv) E Cell = E OPSn2+ /Sn4+ + E RPI2 /I– = –0.1 + 0.54 = + 0.44 V (iv) Is spontaneous change. Ex. 16 Predict whether the following reaction can occur under standard conditions or not. Sn2+ (aq) + Br2 () Sn4+ (aq) + 2Br–(aq) Given Sol. E°cell = E°Sn4+/Sn2+ = + 0.15, E°Br /Br– = 1.06V. 2 E°cathode – E °anode = 1.06 – 0.15 = 0.91V.. Since, E cell comes out be positive, this means that the reaction can occur.. Ex. 17 Co3+ + e–  Co2+ E° = + 1.82V Given that, 2H2O  O2 + 4H+ + 4e– ; E° = –1.23V. Explain why Co3+ is not stable in aqueous solutions. Sol. The Ecell can be calculated as follows : Add : 4 [Co+3 + e–  Co2+] ; E° = + 1.82V 2H2O  O2 + 4H+ + 4e– ; E° = – 1.23 V. 4Co + 2H2O  4Co + 4H + O2 ; E° = 1.82 – 1.23 = + 0.59V. 3+ 2+ + Since E°cell is positive, the cell reaction is spontaneous. This means that Co3+ ions will take part in the reaction. Therefore, Co3+ is not stable. Ex. 18 The measured e.m.f. at 25°C for the cell reaction, Zn (s) + Cu2+ (1.0M)  Cu (s) + Zn2+ (0.1 M) is 1.3 volt Calculate E° for the cell reaction. Sol. Using Nernst equation (at 298 K), E cell = E °cell – Here 0.0591V [Zn 2+ (aq)] log 2 [Cu 2+ (aq)] Ecell = 1.3 V, [Cu2+ (aq)] = 1.0 M, [Zn2+ (aq)] = 0.1 M, Ecell = ? Substituting the values, ° 1.3 V = E cell – 0.0591V 0.1 log 2 1.0 1.3 V = E°cell – 0.02955 V log 10–1 1.3 V = E°cell + 0.02955 V log 10 E°cell = 1.3 V – 0.02955 V = 1.27 V Ex. 19 A Solution of copper (II) sulphate is electrolysed between copper electrodes by a current of 10.0 amperes passing for one hour. What changes occur at the electrodes and in the solution ? ELECTROCHEMISTRY Sol. According to Faraday's first law of electrolysis : The reaction at cathode : Cu2+ + 63.5 2e–  Cu 2 × 96500 C The quantity of charge passed = I × t = (10 amp) × (60 × 60s) = 36000 C. 2 × 96500 C of charge deposit copper = 63.5 g 36000 C of charge deposit copper = (63.5 g) × (36000 C) = 11.84 g (2 × 96500 C) Thus, 11.84 g of copper will dissolve from the anode and the same amount from the solution will get deposited on the cathode. The concentration of the solution will remain unchanged. Ex. 20 Calculate the equilibrium constant for the reaction at 298 K Zn (s) + Cu2+ (aq) ‡ˆ ˆˆ ˆ† ˆˆ Zn2+(aq) + Cu (s) ° Given E °Zn 2+ /Zn = –0.76 V and E Cu 2+ /Cu = +0.34 V Sol. We know that, log Kc = nE ocell 0.0591 ° E °cell =  E Cathode – E °Anode  = [(+0.34 V) – (– 0.76 V)] = 1.10 V, n =2, log K c =  2 × (1.10 V) = 37.29 (0.0591 V) Kc = Antilog 37.29 = 1.95 × 1037 Ex. 21 Calculate the cell e.m.f. and G for the cell reaction at 298 K for the cell. Zn (s) |Zn2+ (0.0004M) ||Cd2+ (0.2M) | Cd(s) o Given E oZn 2+ / Zn = – 0.763 V ; E Cd +2 Cd = – 0.403 V at 298 K. F = 96500 C mol–1. Sol. Step I. Calculation of cell e.m.f. According to Nernst equation E = Eo – 0.0591 [Zn 2+ (aq)] log n [Cd 2+ (aq)] o o Eocell = E (Cd 2+ /Cd) – E (Zn 2+ / Zn) = (–0.403) – (–0.763) = 0.36 V [Zn+ (aq)] = 0.0004 M, [Cd2+ (aq)] = 0.2 M, n = 2 E = (0.36) – (0.0591 V) 0.0004 log 2 0.2 = 0.36 – (0.0591 V) × (–2.69990) 2 = 0.36 V + 0.08 = 0.44 V Step II. Calculation of G  G = –nFEcell Ecell = 0.44 V, n = 2 mol, F = 96500 C mol–1 G = – (2 mol) × (96500 C mol–1) × (0.44 V) = – 84920 CV = – 84920 J Ex.22 Calculate G° for the reaction : Cu2+ (aq) + Fe (s) ‡ˆ ˆˆ ˆ† ˆˆ Sol. The cell reactions are : Fe2+ (aq) + Cu (s) . Given that E°Cu2+/Cu= + 0.34 V,, E Fe (s) Fe2+ (aq) + 2e– ............ At (anode) Cu2+ (aq) + 2e–  Cu (s) ............. At (cathode) We know that : E cell  Fe 2 Fe = – 0.44 V Gh° = –nF E cell ; n = 2 mol, ° ° =  E (Cu 2+ / Cu) – E (Fe 2+ / Fe)  = (+ 0.34 V) – (–0.44 V) = + 0.78 V F = 96500 C mol–1  G° = – nF E cell = –(2mol) × (96500 C mol–1) × (+0.78 V) = – 150540 CV = –150540 J ( 1 CV = 1J) Ex. 23 How will show that Faraday's second law of electrolysis is simply corollary of the first law. Sol. Accroding to Faraday's first law of electrolysis. w=Z× Q If same quantity of electricity is passed through two electolytes, i.e., Q1 = Q2 = Q, then In case of first electrolyte, w1 = Z1 × Q and In case of second electrolyte, On dividing , w2 = Z2 × Q w1 Z1 E1 / 96500 E1 = = = w 2 Z2 E 2 / 96500 E 2 where E1 and E2 are their equivalent masses. Ex. 24 Iron does not rust even if zinc coating on its surface is broken but the same is not true when coating is of tin. Sol. The zinc is more reactive than iron. This means that if a crack appears on the surface of iron coated with zinc even then zinc will take part in the redox reaction and not iron. In other words, zinc will be corroded in preference to iron. But same is not in the case with tin. it is less reactive than iron. If a crack appears on the surface of iron coated with tin, then iron will take part in the redox reaction and not tin. Therefore, iron will be corroded under these circumstances.