Electrochemistry Solved Problems

ELECTROCHEMISTRY SOLVED PROBLEMS Q.1 Calcualate the potential of hydrogen electrode in contact with a solution whose pH is 10. Sol. If pH of solution is 10, that means [H+] = 10–10M According to Nernst equation EH / H  E0  H / H2 2  0.0591  1  log    n H   1  EH / H = 0 – 0.0591 log   = 0 – 0.0591 × 10 = 0 – 0.591 V 2 1  10 10  Q.2 Calculate the emf of the cell in which the following reaction takes place: Ni(s) + 2Ag+(0.002 M)  Ni2+ (0.160 M) + 2Ag(s)  Given that Ecell = 1.05 V. Sol. The oxidation and reduction half reactions are: Ni(s)  Ni2+(aq) + 2e– (At anode) 2Ag+ + 2e–  2Ag Ni(s) + 2Ag+(aq)  Ni2+(aq) + 2Ag(s) According to Nernst equation  – Ecell = Ecell [ Anode] 0.0591 log [Cathode] n   Ecell = [E Ag / Ag  ENi2 / Ni ]  = 1.05 – Ecell = 1.05 – Q.3 0.0591 [Ni2 ] log 2 [ Ag ]2 1.160 0.0591 0.0591 log log(4 × 104) 2 = 1.05 – (0.002) 2 2 0.0591 × 4.6020 = 0.9142 V.. 2 The molar conductivity of 0.025 molL–1 methanoic acid is 46.1 cm2 mol–1. Calculate its degree 0 of dissociation and dissociation constant. Give (H ) = 349.6 S cm 2 mol –1 and 0 (HCOO ) Sol. = 54.6 S cm2 mol–1. (I) Calculation of degree of dissociation () of methanoic acid Cm = 46.15 S cm2 mol–1 0m (HCOOH) = 0m (HCOO–) + 0m (H+) = 54.6 + 349.6 = 404.2 S cm2 mol–1 ELECTROCHEMISTRY = Cm 0m = 2 1 46.15 Scm mol = 0.1140 404 .2 S cm 2 mol 1 (II) Calculation of dissociation constant (KC) HCOOH(aq) HCOO–(aq) + H–(aq) C O O C(1 – ) C C   K = [HCOO ][H ] [HCOOH] K = Q.4 Sol. Sol. Q.6 C 2 1  0.025  (0.1140 )2 = 3.67 × 10–4 molL– (1  0.1140 ) If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire ? Charge = I t Charge =(0.5 A) × (2 × 60 × 60 S) = 3600 AS = 3600 C No. of electorns flowig through the wire tby passing a charge of one faraday (96500 C) = 6.022 × 1023 electrons  No. of electrons flowing through the wire by passing charge of 3600 C = 6.022 × 1023 × Q.5 = 3600C = 2.246 × 1022 electrons 96500 C Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during reacharging. During recharging the cell is operated like an electrolytic cell i.e., electrical energy is supplied from some externa sources. Electrode reactions are reverse of those during discharging. PbSO4(s) + 2e–  Pb(s) + SO42– (aq) Reduction PbSO4(s) + 2H2O(s)  PbO2(s) + SO42–(aq) + 4H+ (aq) + 2e– Oxidation —————————————————————— 2PbSO4(s) + 2H2O(s)  Pb(s) + PBO2(s) + 4H+ (aq) + 2SO42– (aq) Depict the galvanic cell in which the reaction Zn(s) + 2Ag+ (aq)  Zn2+ (aq) + 2Ag(s) takes place. Further show : (i) Which of the electrodes is negatively charged ? (ii) The carrier of the current in the cell. (iii) Individual reactions at each electrode. Reduction + Sol. Zn(s) + 2Ag (aq) Zn2+ (aq) + 2Ag(s) Oxidation Zn(s) (i) (ii) (iii) | Zn2+ (aq) Ag+ (aq) | Ag(s) Daniel cell. Anode of Zn with negative charge ; cathode of Ag with positive charge. Electonrs wil flow from Zn in Cu inouter circuit and current will flow from Cu to Zn. Cathode : Zn  Zn2+ + 2e– ; Anode : 2AG+ + 2e–  Ag(s) Power by: VISIONet Info Solution Pvt. Ltd Website : www.edubull.com Mob no. : +91-9350679141 ELECTROCHEMISTRY Q.7 Calculate the standard cell potentials of galvanic cells in which the following reactions takes place : (i) 2Cr(s) + 3Cd2+(aq)  2Cr3+ (aq) + 3Cd(s) (ii) Fe2+ (sq) + Ag+ (aq) ® Fe3+ (qq) + Ag(s) Calculate the rGº and equilibrium constant of the reactions. Sol. (i) 2Cr  2Cr3+ + 6e–, (Oxidation) 3Cd2 + 6e–  3Cd, (Reduction) Eocell = Eocathode = Eoanode o o = ECd2 / Cd – ECr 3 / Cr = – 0.40 – (– 0.74) = 0.34 V Six electrons (n = 6) are used in redox charge rGº = – nEºF = –6 × 0.34 × 96500 J = – 196860 J or rGº = –196.86 kJ mol–1 Also rGº = – 2.303 × 8.314 × 298 log K K = antilog 34.5014 = 3.173 × 1034 (ii) Fe2+  Fe3+ + e–, (Oxidation) Ag+ + e–  Ag+ (Redution) Eocell = EoCathode – EoAnode o o = E Ag / Ag – EFe3 / Fe 2 = 0.80 – 0.77 = 0.03 V Also rGº = – nFEº = – 1 × 96500 × 0.03 or rGº = – 2895 J mol–1 Also rGº = – 2.303 RT log K –2895 = – 2.303 × 8.314 × 298 log K K = antilog 0.5074 = 3.22 Q.8 Write the Nernst equation and emf of the follownig cell at 298 K : (i) Mg(s) | Mg2+ (0.001 M)|| Cu2+ (0.0001 M) | Cu(s) (ii) Fe(s) | Fe2+ (0.001 M) || H+ (1 M)| H2(g) (g) (1 bar) Pt(s) Sol. (i) Ecell = (EoCathode – Eoanode) – Ecell = (0.34 – (– 2.36)) – [Mg2 ] 0.059 log n [Cu2 ] 0.059 0.001 log = 2.68 V 2 0.0001 (ii) Ecell = (Eocathode – Eoanode) = [Fe 2 ][pH2 ] 0.059 log n [H ]2 Ecell = (0 – (– 0.44)) – 0.001 0.059 log = 0.5285 V (1)2 2 ELECTROCHEMISTRY Q.9 Conductivity of 0.0241 M acetic acid is 7.896 × 10–5 S cm–1. Calculate its molar conductivity. If 0m for acetic acid is 390.5 S cm2 mol–1, what is its dissociation constant? Sol. m = K  1000 7.896  10 5  1000 = = 32.76 S cm2mol–1 C 0.00241 Also  m 0m  32.76 –2 390.5 = 0.084 = 8.4 × 10 2 2 Ka = C  0.00241  (0.084 ) = 1.86 × 10–5 (1   ) (1  0.084 ) Q.10 How much electricity in terms of Faraday ios required to produce (i) 20.0 g of Ca from molten CaCl2 ? (ii) 40.0 g of Al from molten Al2O3 ? Sol. Ca2+ 2Cl– (i) CaCl2 Ca2+ + 2e   Ca 2mol 1mol (1 mole of calcium = 40 cm)  40 g of calcium requires charge = 2F  20 g of calcium requires charge = = 2F  20 40 2  96500  20 = 96500 C 40 (ii) 2Al3+ + 6e   2Al, 6F 227g To produce 2 × 27 g Al, Electrictiy required = 6F  To produce 40 g Al, electricity required = = Q.11 6F × 40 2  27 120 = 4.44 F 27 How much electricity is required in coloumb for the oxidation of (i) 1 mol of H2O to O2 ? (ii) 1 mol of FeO to Fe2O3 ? Sol. 1 O + 2H+ + 2e– 2 2 Thus 1 mol of water will require 2 × 96500C = 193000C (i) The reaction involved is H2O  (ii) Fe2+  Fe3+ e– 1 O  Fe2O3 2 2 So 1 mol of FeO will require 96500 C. or 2FeO + ELECTROCHEMISTRY Q.12 Using the standard electrode potentials given in the Table, predict if the reaction between the following is feasible: (a) Fe3+(aq) and I–(aq) (b) Fe3+(aq) and Br–(aq) (c) Br2(aq) and Fe2+(aq) Electrode Eo Fe3 / Fe 2 Eº Sol. (a) 0.77 V Eo o EBr 2 / Br 0.54 V 1.09 V I2 / I 2Fe3+ + 2e– 2 I–  2Fe2+  I2 + 2e– ____________________________________ 2Fe3+ + 2I–  Fe2+ + I2 Thus two half cells are Fe3/Fe2+ and I2/I– Eºcell = Eºcathode – Eºanode 0 0 = E Ag3 / Ag2 – EI / I = 0.77 – 0.54 = 0.23 V 2 This reaction is feasible as Eºcell is positive (b) 2Fe3+ + 2e– 2Br– + 2e–  2Fe2+  Br2 ____________________________________ Br– + Fe3+  Fe2+ + 1 Br 2 2 –––––––––––––––––––––––––––––––––––––––––––––– Eºcell = Eºcathode – Eºanode 0 0 = E Ag3 / Ag2 – EBr2 / Br = 0.77 – 1.09 = –0.32 V The reaction is not feasible as Eºcell is negative. (c) Br2 + 2e–  2Br– 2Fe2+ 2Fe3+ + 2e– ____________________________________ Br2 + 2Fe2+  2Br– + 2Fe3+ –––––––––––––––––––––––––––––––––––––––––––––– Thus two half cells are Br2/Br– and Fe3+/Fe2+ Eºcell = Eºcathode – Eºanode 0 0 = EBr2 / Br  – E Ag3 / Ag2 = 1.09 – 0.77 = 0.32 V The reaction is feasible as Eºcell is positive. So the reaction Br2 + 2Fe2+  2Br– + 2Fe3+ ELECTROCHEMISTRY Q.13 In the electrolysis of acidulated water, it is desired to obtain H2 at the rate of 1cc per second at NTP condition. What should be the current passed ? 2H+ + 2e–  H2 Sol. 2F 1 mole 22400 cc at NTP  22400 cc of H2 at NTP requires 2F i.e., 2 × 96500 C  1 cc of H2 at NTP requires Now, Q=I×t I= Q.14 2  96500 × 1 = 8.616 C 22400 Q 8.616 = = 8.616 ampere t 1s Consider the following diagram of mercury cell (a) Which substance acts as cathode and which acts as anode ? Anode (b) Mention the electrolyte used in the cell (c) (d) What is the cell potential of this cell ? Why does the voltage of mercury cell remain constant throughout in life ? Sol. (a) Anode – Zinc container, Cathode – Carbon rod Greaket Separator Cathode Anode cap Cell can (b) A paste of HgO + KOH (c) 1.35 V (d) Because overall cell reaction does not involve any ion whose concentration can change Zn(s) + HgO(s)  ZnO(s) + Hg(s) Q.15 At what pH will hydrogen electrode at 298 K show an electrode potential of – 0.118 V, when H2 gas is bubbled at 1 atm. pressure. Sol. The electrode reaction involved is: H+(aq) + e–  EH / H  E0  H / H2 2 – 0.118 = 0 – log 1 [H ] = 1 H 2 2(g)  1 0.0591 log [H ] 1 0.118 0.0591 – log[H+] = 1.996  2 pH = 2 0.0591 1 log  n [H ]