CHAPTER 10 ELECTROCHEMISTRY TEXT BOOK EXERCISE Q1. Multiple choice questions. (i) The cathode reaction in the electrolysis of dill. H2SO4 with Pt electrode is (a) Reduction (b) Oxidation (c) Both oxidation and reduction (d) Neither oxidation nor reduction (ii) Which of the following statement is correct about galvanic cell? (a) Anode is negatively charged (b) Reduction occurs at anode (c) Cathode is positively charged (d) Reduction occurs at cathode (iii) Stronger the oxidizing agent, greater is the (a) Oxidation potential (b) Reduction potential (c) Redox potential (d) EMF of cell (iv) If the slat bridge is not used between two half cells, then the voltage (a) Decrease rapidly (b) Decrease slowly (c) Does not change (d) Drops to zero (v) If a strip of Cu metal is placed in a solution of FeSO4. (a) Cu will be precipitated out (b) Fe is precipitated out (c) Cu and Fe both dissolve 1

(d) No reaction takes place Ans. i) a ii) d iii) b iv) d v) d Q2. Fill in the blanks with suitable words. (i) The oxidation number of O atom is in OF2 and is in H2O2. (ii) Conductivity of metallic conductors is due to the flow of while that of electrolytes is due to flow of. (iii) Reaction taking place at the is termed as oxidation, and at the is called reduction. (iv) is setup when a metal is dipped in its own ions. (v) Cu metal the Cu-cathode when electrolysis is performed for CuSO4 solution with Cu-cathodes. (vi) The reduction potential of Zn is volts and its oxidation potential is volts. (vii) In a fuel cell react together in the presence of. Ans. i) +2, -1 ii) electrons. Ions iii) anode, cathode iv) Equilibrium v) deposits on vi) -0.76, +0.76 vii) H2 and O2, Pt catalyst Q3. Indicate TURE or FALSE as the case may be. (i) In electrolytic conduction electrons flow through the electrolyte. (ii) In the process of electrolysis, the electrons in the external circuit flow from cathode to anode. (iii) Sugar is a non-electrolyte in solid form and when dissolved in water will allow the passage of an electric current. (iv) A metal will only allow the passage of an electric current when it is in cold state. (v) The electrolyte products of aqueous copper (II) chloride solution 2

are copper and chlorine. (vi) Zinc can displace iron form its solution. (vii) SHE cats as cathode when connected with Cu electrode. (viii) A voltaic cell produces electrical energy at the expense of chemical energy. (ix) Lead storage battery is not reversible cell. (x) Cr changes its oxidation number when K2 Cr2 O7 is reacted with HCI. Ans. i) False ii) False iii) False iv) False v) True vi) True vii) False viii) True ix) False x) True Q4. Describe the electrolysis of molten sodium chloride, and a concentrated solution of sodium chloride. See Section 10.2.4 Q5. What is the difference between single electrode potential and standard electrode potential? How can it be measured? Give its importance. Q6. Outline the important applications of electrolysis. Also write the electrochemical reactions involved therein. Discuss the electrolysis of Cu SO4 using Cu electrode and AgNO3 solutions using Ag electrode. See Section 10.2.2 Q7. Describe the construction and working of standard hydrogen electrode. See Section 10.3.1 Q8. Is the reaction Fe3+ + Ag EMBED Equation.DSMT4 Fe2+ + Ag+ spontaneous? If not, write spontaneous reaction involving these species. Solution Fe3+ + Ago EMBED Equation.DSMT4 Fe2+ + Ag+ 3

In this reaction, Fe is reduced while Ag is oxidized. Therefore, Fe+3 will act as cathode while Ago as anode. Thus, emf of the cell will be Eocell =Eoox + Eored Eocell = - 0.7994 + (- 0.44) Eocell = - 0.7994-0.44 Eocell = - 1.2394 Since emf of cell is negative, therefore, the cell-reaction is nonspontaneous. But if the electrodes are reversed, the cell-reaction becomes spontaneous i.e. Fe3+ + Ago EMBED Equation.DSMT4 Fe3+ + Ago Q9. Explain the difference between (a) Ionization and electrolysis IONIZATION ELECTROLYSIS 1 The process in which ionic compounds when fused or dissolved in water split up into charged particles called ions. 1 The process in which electricity is used to carry out a non-spontaneous reaction is called electrolysis. 2 Electrodes are not needed 2 Electrodes are required 3 Electricity is not needed 3 Electricity is required 4 Since there are no electrodes, therefore, ions do not move towards electrodes 4 Ions moves towards their respective electrodes 5 After ionization, ions are not discharged 5 Ions are discharged at electrodes to give neutral products. (b) Electrolytic and Voltaic cell See Section 10.2.2 and 10.2.5 (c) Conduction through metals and molten electrolytes CONDUCTION THROUGH METALS CONDUCTION THROUGH MOLTEN ELECTROLYTE 1 Electrical conduction takes place due 4

to free electrons. 1 Electrical conduction takes place due to ions 2 There in no need to convert metal into molten state. 2 Electrolyte must be converted into molten state for electrical conduction 3 In this case, conductance decreases with increase in temperature. 3 In this case, conductance increase with increase in temperature. 4 No chemical reaction occurs during conduction. 4 Chemical reaction occurs take place during conduction. 5 Chemical composition of metal is not changed during conduction and no new substance are produced. 5 Since chemical reactions occur, therefore new substances are produced. 6 Example: All metals are conductions. e.g. Fe, Pb etc. 6 Example: Molten salt e.g. NaCI(l) or their aqueous solutions, acids, bases etc. Q10. Describe a galvanic cell explaining the function of electrodes and salt bridge. See Section 10.2.5 Q11. Write comprehensive notes on Spontaneity of oxidation-reduction reactions. See Section 10.4.1 (b) Electrolytic conduction See Section 10.2 (c) Alkaline, silver oxide and Nickle-Cadmium batteries, fuel cells. See Section 10.5 (d) Lead accumulator, its desirable and undesirable features. See Section 10.5.1 Q12. Will the reaction be spontaneous for the following set of half reactions? What will be the value of Ecell? (i) Cr3+(aq) + 3e- EMBED Equation.DSMT4 Cr(s) (ii) MnO2(s) + 4H+ + 2e- EMBED Equation.DSMT4 Mn2+(aq) + 2H2O(l) 5

Standard reduction potential for reaction (i) = - 0.7 and for the reaction (ii) = + 1.28 V. In reaction (i) Cr is reduced from +3 to +2. In reaction (ii) Mn is also reduction reactions; hence these reactions are not possible in these forms. However, if reaction (i) is reversed so that Cr is oxidized then the reaction becomes spontaneous and its emf can be calculated as (i) Cr(s) EMBED Equation.DSMT4 Cr3+(aq) + 3e- Eoox = + 0.74V (ii) MnO2(s) + 4H+ + 2e- EMBED Equation.DSMT4 Mn2+(aq) + 2H2O(l) Eo(aq) =+1.28 V emf of the cell is given by Eocell =Eooxl + Eored Eocel = +0.74 + 1.28 Eocell =2.02 V Q13. Explain the following with reasons A porous plate or a salt bridge is not required in lead storage cell. A salt bridge has two main functions It joins solution of two half cells and thus complete the circuit. It maintains electrical neutrality of the two half cells as ions can pass through it. In lead storage battery, both cathode and anode are dipped in the same solution. Therefore, excess positive or negative ions are not produced in the solution. Hence, there is no need of salt bridge. The standard oxidation potential of Zn is 0.76 V and its reduction potential is 0.76 V. According to the law of conservation of energy, energy can neither be created nor destroyed. Therefore, if standard oxidation potential of Zn is 0.76 V, then its potential for reverse process, i.e. standard reduction potential will also be same but with positive sign. Thus 6

Zn EMBED Equation.DSMT4 Zn2+ + 2e- Eoox =0.76 V Zn2+ + 2e- EMBED Equation.DSMT4 Zn Eoox =0.76 V (C) Na and K can displace hydrogen from acids but Pt, Pd and Cu cannot. Greater the value of reduction potential, Lesser is the ability to loose electron to from positive ion, Hence weaker is its tendency to displace H2-. Metals like Pt, Pb, and Cu have high positive value of reduction potential. Thus these do not liberate H2. Metals like Na and K have negative values of reduction potential. Thus, these can liberate H2. 2Na +2HCI EMBED Equation.DSMT4 2NaCI + H2 2K +2HCI EMBED Equation.DSMT4 2KCI + H2 The equilibrium is set up between metal atoms of electrode and ions of metal in a cell. When a metal electrode is dipped into the solution of its own ion. there may be two tendencies Metal atom from electrode leaves the electron on metal an goes into solution. this is oxidation process M EMBED Equation.DSMT4 M+ + e- Metal ions in solution may take up electrons form the metal electrode and deposit as atom on electrode. This is reduction process. M+ e- EMBED Equation.DSMT4 M+ At last, a dynamic equilibrium is established due to same rate of two processes. Thus no further potential difference is developed. (e) A salt bridge maintains the electrical neutrality in the cell. Two half cells are electrically connected by a salt bridge. Consider a Zn-Cu cell During reactions of this cell, Zn half cell continuously loose 7

electrons. Thus, in this positive charge is increasing. Zn EMBED Equation.DSMT4 Zn2+ + 2e- While, Cu half cell continuously receive electrons, thus it goes on collecting negative charge. Cu2+ + 2e- EMBED Equation.DSMT4 Cu Collection of positive charge in Zn electrode half cell and collection of negative charge in Cu half cell would stop the reaction. Salt bridge prevents the net accumulation of charges in either beaker. Thus form negative Cu half cell, negative ions diffuse through the salt bridge into the positive Zn half cell. In this way, salt bridge maintains the two solution, electrically neutral. (f) lead accumulator is a chargeable battery. See Section 10.5.1 (g) Impure Cu can be purified by electrolytic process. Impure Cu can be made pure in an electrolytic cell. Thick sheets of impure copper are made anode, while, thin sheets of pure copper are made cathode in the cell. These sheets are placed in an electrolytic solution of CuSO4. When current is passed through the cell, Cu from anode is oxidized to Cu2+ ions, which go into the solution. from the solution, Cu2+ ions are reduced to metallic Cu and deposits as pure Cu on cathode. In this way, impure sheets of Cu (anode) become this, while pure sheets of pure Cu (cathode) become thick. The reactions in the cell are At Anode (oxidation) Cu EMBED Equation.DSMT4 Cu2+ + 2e- At Cathode (reduction) Cu2+ + 2e- EMBED Equation.DSMT4 Cu 8

Thus, there is no net reaction in the cell. However, the net result is the purification of Cu. (h) S.H.E. acts as anode when connected with the Cu electrode but as cathode with Zn electrode. See Section 10.3.2 NUMERICAL PROBLEMS (Exercise) Q14. (c) Calculate the oxidation number of Chromium in the following. CrCl3 K2CrO4 Oxidation number of Cl=-1 Oxidation number of K = + 1 Oxidation number of Cr=x Oxidation number of Cr= - 2 Oxidation number of Cr Oxidation number of Cr=x can be calculated as Thus For CrCl3 For K2CrO4 x + 3( - 1) =0 2(+1) + x +4(-2)=0 x 3 =0 x 6 =0 or x =+3 x =+6 K2Cr2O7 CrO3 Oxidation number of K=+1 Oxidation number of O= -2 Oxidation number of O= -2 Oxidation number of Cr=x 9

Thus Thus Oxidation number of Cr=x For K2Cr2O7 For CrO3 2(+1)+2x + 7(-2) =0 x +3(-2)= 0 2x 12 =0 x 6 =0 Or x=+12/2 =+6 Or x =+6 Cr2O3 Cr2O72- Oxidation number of O =-2 Oxidation number of O =-2 Oxidation number of Cr = x Oxidation number of Cr = x Thus Thus For Or2O3 For Cr2O72-2x+3(-2)=0 2x +7(-2) = -2 2x 6 = 0 2x 14= - 2 +14 Or x= +6/2 =+3 Or x = + 12/2 =+6 Cr2(SO4)3 Oxidation number of S = +16 Oxidation number of O = -2 Oxidation number of Cr = x Thus For Cr2(SO4)3 2x +3[(+6)+4(-2)]=0 10

2x 6 =0 Or x=6/2 =+3 (d) Calculate the oxidation numbers of the elements underlined in the following compounds Na3 PO4 Na2CO3 Oxidation number of Na = +1 Oxidation number of O = -2 Oxidation number of Na = +1 Oxidation number of P = x Oxidation number of C = x Thus For Na3 PO4 For Na2CO3 3(+1) + x + 4(-2) =0 2(+1) + x + 3 (-2) = 0 x 5 =0 x 4 =0 Or x = +5 Or x =+4 Cr2(SO4)3 Ca(ClO3)2 Oxidation number of Cr = +3 Oxidation number of Ca = + Oxidation number of O = - 2 Oxidation number of O = - 2 Oxidation number of S = x Oxidation number of S = x Thus Thus Thus For Cr2(SO4)3 For Ca(ClO3)2 2(+3) +3 [ (x)+4(-2)] =0 (+2) +2 [ (x)+3(-2)] =0 11

3x 18 =0 2 + 2x -12 =0 Or x = 18/3 = +6 Or x=10/2 = +5 K2 MnO4 HNO3 Oxidation number of K = +1 Oxidation number of H = +1 Oxidation number of O = -2 Oxidation number of O = -2 Oxidation number of Mn = x Oxidation number of N = x Thus Thus For K2 MnO4 For HNO3 2(+1) + x + 3(-2) = 0 (+1) + x +3(-2) =0 x 6 =0 x 5 =0 Or x = +6 Or x = +15 HPOs Oxidation number of H = +1 Oxidation number of O = -2 Oxidation number of P = x Thus For HPOs (+1) + x + 3(-2) =0 x - 5 =0 Or x = +5 Try Yourself H 2O2, Ca(OCl)2, NalO2, Zn(OH)2, H3PO4, Q15. (b) Balance the following equations by oxidation number 12

method. PROBLEM HNO3 + Hl EMBED Equation.DSMT4 NO + l2 + H2O Identify the elements, which undergo a change in oxidation number and write their oxidation numbers over the symbols. +1+5 2(-2) +1-1 +2 _-2 o HNO3 + Hl EMBED Equation.DSMT4 NO + l2 + H2O Determine the no. of electrons gained and lost and equate them. gain of 3e- x 1 =3 e- (reduction ) +5-1 +2 o HNO3 + Hl EMBED Equation.DSMT4 NO + l2 + H2O lose of l e- x 3 = 3 e- (oxidation) Balance loss and gain of electrons by multiplying Hl by 3. HNO3 + 3Hl EMBED Equation.DSMT4 NO + l2 + H2O Balance the rest of equation by inspection method. 2HNO3 + 6Hl EMBED Equation.DSMT4 2NO + 3l2 + 4H2O PROBLEM Br2 + NaOH EMBED Equation.DSMT4 NaBrO3 + NaBr + H2O Identify the elements which undergo a change in oxidation number and write their oxidation number over the symbols. o +1-2 + 1 +1 +5 3(-2) +1-1 13

Br2 + NaOH EMBED Equation.DSMT4 NaBrO3 + NaBr + H2O Since Br2 is involved both in oxidation and reduction, therefore, we shall write the Br=twice. Then determine the no. of electrons gained and lost and equate them. gain of 2(l) e- x 5 = 10e- (reduction ) o +1 +5 3(-2) +1-1 Br2 + Br2 + NaOH EMBED Equation.DSMT4 NaBrO3 + NaBr + H2O loss of 2(5) e- x 1 = 10 e- (reduction) Balance loss and gain of electrons by multiplying Br2 by 5, in which oxidation occurs. 5Br2 + Br2 + NaOH EMBED Equation.DSMT4 NaBrO3 + NaBr +H2O Balance the rest of equation by inspection method. 5Br2 + Br2 +12 NaOH EMBED Equation.DSMT4 2NaBrO3 + 10NaBr + 6H2O Or 6Br2 +12 NaOH EMBED Equation.DSMT4 2NaBrO3 + 10NaBr + 6H2O Or 3Br2 +6 NaOH EMBED Equation.DSMT4 NaBrO3 + 5NaBr + 3H2O PROBLEM Zn + HNO3 EMBED Equation.DSMT4 Zn(NO3) + NO +H2O Identify the elements, which undergo a change in oxidation number and write their oxidation number over the symbols. 0 +1+5 3(-2) +2+5 +2-2 Zn + HNO3 EMBED Equation.DSMT4 Zn(NO3) + NO +H2O In this eq. N is reduced from +5 in HNO3 to +2 in NO. But it s 14

oxidation state is not change in Cu(NO3)2. therefore, write HNO3 twice and determine the number of electrons gained and lost and equate them. gain of 3 e- x 2 = 6 e- (reduction ) 0 +5 +2 +5 +2-2 HNO3 +Zn + HNO3 EMBED Equation.DSMT4 Zn(NO3) + NO +H2O loss of 2 e- x 3 = 6 e- (oxidation ) Use the multiplier obtained above to balance loss and gain of electrons. Thus Cu is multiplied by 3 and HNO3 by 2. HNO3 +3Zn + 2HNO3 EMBED Equation.DSMT4 Zn (NO3) + NO +H2O Balance the rest of equation by inspection method. 6HNO3 +3Zn + 2HNO3 EMBED Equation.DSMT4 3Zn(NO3) + 2NO +4H2O Or 3Zn + 8HNO3 EMBED Equation.DSMT4 3Zn(NO3) + 2NO +4H2O PROBLEM +4 2(-2) + -1 +2 2(-1) 0 MnO2 HCI EMBED Equation.DSMT4 MnCl2 + H2O + Cl2 Identify eq. Cl is reduced from 1 in HCl to 0 in Cl2. But its oxidation state is not changed in MnCl2. therefore, write HCl twice and determine the no. of electrons gained and lost and equate them. gain of 2 e- x 1 = 2 e- (reduction ) 15

+4-1 +2 2(-1) 0 HCI+MnO2 HCI EMBED Equation.DSMT4 MnCl2 + H2O + Cl2 loss of 1 e- x 2 = 2 e- (oxidation ) Use the multiplier obtained above to balance loss and gain of electrons. HCI+MnO2 2HCI EMBED Equation.DSMT4 MnCl2 + H2O + Cl2 Balance the rest of equation by inspection method. 2HCI+MnO2 2HCI EMBED Equation.DSMT4 MnCl2 + 2H2O + Cl2 Or MnO2 + 4HCI EMBED Equation.DSMT4 MnCl2 + 2H2O + Cl2 PROBLEM FeSO4 + K2Cr2O7 + H2SO4 EMBED Equation.DSMT4 Fe(SO4) + Cr2(SO4)3 + K2SO4 + H2O Identify the elements, which undergo a change in oxidation number and write their oxidation numbers over the symbols. +2 +6 +3 +3 FeSO4 + K2Cr2O7 + H2SO4 EMBED Equation.DSMT4 Fe(SO4) + Cr2(SO4)3 + K2SO4 + H2O Determine the no of electrons gained and lost and equate them. gain of 2(3) e- x 1 = 6 e- (reduction ) +2 +6 +3 +3 FeSO4 + K2Cr2O7 + H2SO4 EMBED Equation.DSMT4 Fe(SO4) + Cr2(SO4)3 + K2SO4 + H2O 16

loss of 1(2) e- x 3 = 6 e- (oxidation ) Use the multiplier obtained above to balance loss and gain of electrons 6FeSO4 + K2Cr2O7 + H2SO4 EMBED Equation.DSMT4 3Fe (SO4) + Cr2 (SO4)3 + K2SO4 + H2O Balance the rest of equation by inspection method. 6FeSO4 + K2Cr2O7 + 7H2SO4 EMBED Equation.DSMT4 3Fe(SO4) + Cr2(SO4)3 + K2SO4 + 7H2O PROBLEM Cu + H2SO4 CuSO4 + SO2 + H2O Identify the elements, which undergo a change in oxidation number and write their oxidation number over the symbols. 0 +6 +2 +4 2((-2) Cu + H2SO4 EMBED Equation.DSMT4 CuSO4 + SO2 + H2O In this equation S is reduced from +6 in H2SO4 to +4 in SO2. But it s oxidation state is not changed in CuSO4. Therefore, write H2SO4 twice and determine the no. of electrons gained and lost equate them. gain of 2 e- x 1 = 2 e- (reduction ) 0 +6 +2 +4 H2SO4+ Cu + H2SO4 EMBED Equation.DSMT4 CuSO4 + SO2 + H2O loss of 2 e- x 1 = 2 e- (oxidation ) Use the multiplier obtained above to balance loss and gain of electrons. H2SO4+ Cu + H2SO4 EMBED Equation.DSMT4 CuSO4 + SO2 + H2O Balance the rest of equation by inspection method. H2SO4+ Cu + H2SO4 EMBED Equation.DSMT4 CuSO4 + SO2 + 17

2H2O Or Cu + 2H2SO4 EMBED Equation.DSMT4 CuSO4 + SO2 + 2H2O PROBLEM H2SO4+ Hl EMBED Equation.DSMT4 SO4 + l2 + H2O Identify the elements, which undergo a change in oxidation number and wirte their oxidation number over the symbols. +6 +1-1 +4 2(-2) 0 H2SO4+ Hl EMBED Equation.DSMT4 SO4 + l2 + H2O Determine the no. of electrons gained and lost and equate them. gain of 2 e- x 1 = 2 e- (reduction ) H2SO4+ Hl EMBED Equation.DSMT4 SO4 + l2 + H2O loss of 2 e- x 1 = 2 e- (oxidation ) Use the multiplier obtained above to balance loss and gain of electrons H2SO4+ 2Hl EMBED Equation.DSMT4 SO4 + l2 + H2O Balance the rest of equation by inspection method. H2SO4+ 2Hl EMBED Equation.DSMT4 SO4 + l2 + 2H2O PROBLEM NaCl + MnO2 + H2SO4 EMBED Equation.DSMT4 Na2SO4 + MnSO4 + H2O + Cl2 Identify the elements, which undergo a change in oxidation number and write their oxidation number over the symbols. +1-1 +4 2(-2) +2 0 18

NaCl + MnO2 + H2SO4 EMBED Equation.DSMT4 Na2SO4 + MnSO4 + H2O + Cl2 Determine the number of electrons gained and lost and equate them. gain of 2 e- x 1 = 2 e- (reduction ) -1 +4 +2 0 NaCl + MnO2 + H2SO4 EMBED Equation.DSMT4 Na2SO4 + MnSO4 + H2O + Cl2 loss of 2 e- x 1 = 2 e- (oxidation ) Use the multiplier obtained above to balance loss and gain of electrons 2NaCl + MnO2 + H2SO4 EMBED Equation.DSMT4 Na2SO4 + MnSO4 + H2O + Cl2 Balance the rest of equation by inspection method. 2NaCl + MnO2 + H2SO4 EMBED Equation.DSMT4 Na2SO4 + MnSO4 + H2O + Cl2 PROBLEM K2Cr2O7 + HCl EMBED Equation.DSMT4 KCl + CrCl3 + H2O + Cl2 Identify the elements, which undergo a change in oxidation number and write their oxidation number over the symbols. +6-1 +3 0 K2Cr2O7 + HCl EMBED Equation.DSMT4 KCl + CrCl3 + H2O + Cl2 In this eq. Cl is oxidation from -1 in HCl to 0 in Cl2. But it s oxidation state is not change in KCl / or CrCl3. therefore, write HCl twice and determine the no. of electrons gained and lost and equate them. gain of 2(3) =6 e- x 1 = 3 e- (reduction ) +6-1 +3 0 19

HCl+ K2Cr2O7 + HCl EMBED Equation.DSMT4 KCl + CrCl3 + H2O + Cl2 loss of 1 e- x 6 = 6 e- (oxidation ) Use the multiplier obtained above to balance loss and gain of electrons HCl+ K2Cr2O7 + 6HCl EMBED Equation.DSMT4 KCl + CrCl3 + H2O + Cl2 Balance the rest of equation by inspection method. 8HCl+ K2Cr2O7 + 6HCl EMBED Equation.DSMT4 2KCl +2CrCl3 + 7H2O + 3Cl2 Or K2Cr2O7 + 14HCl EMBED Equation.DSMT4 2KCl + 2CrCl3 + 7H2O + 3Cl2 Q16. (b) Balance the following equation by ion-election method PROBLEM Sn2+ + Fe3+ EMBED Equation.DSMT4 Sn4+ + Fe2+ Identify the elements, which undergo oxidation and reduction and split up the reaction into oxidation and reduction half reaction. Fe3+ EMBED Equation.DSMT4 Fe2+ Sn2+ EMBED Equation.DSMT4 Sn4+ Write down the number of electrons gained and lost in each half reaction Fe3+ +le- EMBED Equation.DSMT4 Fe2+ (reduction half reaction) (1) 20

Sn2+ EMBED Equation.DSMT4 Sn4+ + 2e- (oxidation half reaction) (2) Equate the total number of electrons gained and lost by multiplying eq. (1) by 2, and then add them. 2Fe3+ +2e- EMBED Equation.DSMT4 2Fe2+ Sn2+ EMBED Equation.DSMT4 Sn4+ + 2e- Sn2+ +2Fe3+ EMBED Equation.DSMT4 Sn4+ +2Fe2+ PROBLEM H+ + Cl- + Cr2O72- EMBED Equation.DSMT4 Cr3++ Cl2 Identify the elements, which undergo oxidation and reduction and split up the reaction into oxidation and reduction half reactions. Cr2O72- EMBED Equation.DSMT4 Cr3+ (reduction half reaction) 2Cl- EMBED Equation.DSMT4 Cl2 (oxidation half reaction) Balance oxygen by adding H2O. Cr2O72- EMBED Equation.DSMT4 Cr3+ + 7H2O 2Cl- EMBED Equation.DSMT4 Cl2 Balance hydrogen by adding H+ ions. 14H+ + Cr2O72- EMBED Equation.DSMT4 2Cr3+ + 7H2O 2Cl- EMBED Equation.DSMT4 Cl2 Write down the number of electrons gained and lost in each half reaction. 14H+ + Cr2O72- + 6e- EMBED Equation.DSMT4 2Cr3+ + 7H2O (1) 2Cl- EMBED Equation.DSMT4 Cl2 +2e- (2) Equate the total number of electrons gained and lost by multiplying eq.(2) by 3. And then add the two half reactions. 14H+ + Cr2O72- + 6e- EMBED Equation.DSMT4 2Cr3+ + 7H2O 21

6Cl- EMBED Equation.DSMT4 3Cl2 + 6e- Cr2O72- +14H+ + 6e- EMBED Equation.DSMT4 2Cr3+ +NO2+ 7H2O PROBLEM (Acidic Media) Cu + NO3-1 + H+ EMBED Equation.DSMT4 Cu+2 + NO2+ + H2O Identify the elements, which undergo oxidation and reduction and split up the reaction into oxidation and reduction half reactions. NO3-1 EMBED Equation.DSMT4 NO2 (reduction half reaction) Cu EMBED Equation.DSMT4 Cu2+ (oxidation half reaction) Balance oxygen by adding H2O. NO3-1 EMBED Equation.DSMT4 NO2 +H2O Cu EMBED Equation.DSMT4 Cu2+ Balance hydrogen by adding H+ ions. 2H+ + NO3-1 EMBED Equation.DSMT4 NO2 Cu EMBED Equation.DSMT4 Cu2+ Write down the number of electrons gained and lost in each half reaction 4H+ + NO3-1 + le- EMBED Equation.DSMT4 NO2 +H2O (1) Cu EMBED Equation.DSMT4 Cu2+ + 2e- (2) Equate the total number of electrons gained and lost by multiplying eq. (1) by 2. And then add the two half reactions. 4H+ + 2NO3-1 + 2e- EMBED Equation.DSMT4 2NO2 +2H2O Cu EMBED Equation.DSMT4 Cu2+ + 2e- 2NO3-1 +4H+ + Cu EMBED Equation.DSMT4 Cu2+ + 2NO2 22

+2H2O PROBLEM (Acidic Media) Cr2O72- + H3AsO3 EMBED Equation.DSMT4 Cr3 + H3AsO4 Identify the elements, which undergo oxidation and reduction and split up the reaction into oxidation and reduction half reactions. Cr2O72- EMBED Equation.DSMT4 2Cr3+ (reduction half reaction) H3AsO3 EMBED Equation.DSMT4 H3AsO4 (oxidation half reaction ) Balance oxygen by adding H2O. Cr2O72- EMBED Equation.DSMT4 2Cr3 + 7H3O H2O + H3AsO3 EMBED Equation.DSMT4 H3AsO4 Balance hydrogen by adding H+ ions. 14H+ +6e-+ Cr2O72- + EMBED Equation.DSMT4 2Cr3 + 7H3O (1) H2O +3 H3AsO3 EMBED Equation.DSMT4 3 H3AsO3 + 2e- + 2H+ (2) Equate th total number of electrons gained and lost by multiplying eq. (2) by 3. And then add the two half reactions. 14H+ +6e-+ Cr2O72- + EMBED Equation.DSMT4 2Cr3 + 7H3O 3H2O +3 H3AsO3 EMBED Equation.DSMT4 3 H3AsO3 + 6e- + H+ Cr2O72- +8H+ + 3H3AsO3 EMBED Equation.DSMT4 2Cr3 + 3H3AsO4 + 4H2O 23

PROBLEM (Acidic Media) MnO4-+ Cr2O72- EMBED Equation.DSMT4 Mn2+ + CO2 Identify the elements, which undergo oxidation and reduction and split up the reaction into oxidation and reduction half reactions. MnO4- EMBED Equation.DSMT4 Mn2+ (reduction half reaction) Cr2O72- EMBED Equation.DSMT4 2CO2 (oxidation half reaction) Balance oxygen by adding H2O. MnO4- EMBED Equation.DSMT4 Mn2+ + 4H2O Cr2O72- EMBED Equation.DSMT4 2CO2 Balance hydrogen by adding H+ ions. 8H+ +MnO4- EMBED Equation.DSMT4 Mn2+ + 4H2O Cr2O72- EMBED Equation.DSMT4 2CO2 Write down th number of electrons gained and lost in each half reaction 5e-+ 8H+ +MnO4- EMBED Equation.DSMT4 Mn2+ + 4H2O (1) Cr2O72- EMBED Equation.DSMT4 2CO2 +2e-1 (2) Equate th total number of electrons gained and lost by multiplying eq. (2) by 5 and eq. (1) by 2. And then add the two half reactions. 10e-+ 16H+ +2MnO4- EMBED Equation.DSMT4 2Mn2+ + 8H2O + 10CO2 5Cr2O72- EMBED Equation.DSMT4 10CO2 +10e-1 24

5C2O42-+ 16H+ +2MnO4- EMBED Equation.DSMT4 2Mn2+ + 8H2O + 10CO2 PROBLEM (Acidic Media) Fe2+ + C2O72- EMBED Equation.DSMT4 Cr3+ + Fe3+ Identify the elements, which undergo oxidation and reduction and split up the reactions into oxidation and reduction half reactions. C2O72- EMBED Equation.DSMT4 Cr3+ (reduction half reaction) Fe2+ EMBED Equation.DSMT4 Fe3+ (oxidation half reaction) Balance oxygen by adding H2O. C2O72- EMBED Equation.DSMT4 2Cr3+ + 7H2O. Balance hydrogen by adding H+ ions. 14H++ C2O72- EMBED Equation.DSMT4 2Cr3+ + 7H2O. Fe2+ EMBED Equation.DSMT4 Fe3+ Write down th number of electrons gained and lost in each half reaction 14H+ +6e-+ Cr2O72- EMBED Equation.DSMT4 2Cr3 + 7H3O (1) Fe2+ EMBED Equation.DSMT4 Fe3+ + le- (2) Equate the total number of electrons gained and lost by multiplying eq. (2) by 6. And then add the two half reactions. 14H+ +6e-+ Cr2O72- EMBED Equation.DSMT4 2Cr3 + 7H3O 6Fe2+ EMBED Equation.DSMT4 6Fe3+ + 6e- Cr2O72- +14H+ +6e- EMBED Equation.DSMT4 6Fe2++ 2Cr3 + 7H3O 25

PROBLEM (Acidic Media) lo3- + AsO33- EMBED Equation.DSMT4 l- + AsO43- Identify the elements, which undergo oxidation and reduction and split up the reaction into oxidation and reduction and reduction half reactions. lo3- EMBED Equation.DSMT4 l- (reduction half reaction) AsO33- EMBED Equation.DSMT4 AsO43- (oxidation half reaction) Balance oxygen by adding H2O. lo3- EMBED Equation.DSMT4 l- + 3H2O H2O+ AsO33- EMBED Equation.DSMT4 AsO43- Balance oxygen adding H+ ions. 6H+ + lo3- EMBED Equation.DSMT4 l- + 3H2O H2O+ AsO33- EMBED Equation.DSMT4 AsO43- + 2H+ Write down the number of electrons gained and lost in each half reaction 6e- + 6H+ + lo3- EMBED Equation.DSMT4 l- + 3H2O (1) H2O+ AsO33- EMBED Equation.DSMT4 AsO43- + 2H+ + 2e - (2) Equate the total number of electrons gained and lost by multiplying eq. (2) by 3. And then add the two half reactions. 6e- + 6H+ + lo3- EMBED Equation.DSMT4 l- + 3H2O 3H2O+ 3AsO33- EMBED Equation.DSMT4 3AsO43- + 6H+ + 6e - 26

lo3- + 3AsO33 EMBED Equation.DSMT4 3AsO43- +l- PROBLEM (Acidic Media) Cr3+ + BiO3 EMBED Equation.DSMT4 Cr2O72- + Bi3+ Identify the elements, which undergo oxidation and reduction and split up the reaction into oxidation and reduction half reactions. BiO3 EMBED Equation.DSMT4 Bi3+ (reduction half reaction) Cr3+ EMBED Equation.DSMT4 Cr2O72- (oxidation half reaction) Balance oxygen by adding H2O. BiO3 EMBED Equation.DSMT4 Bi3+ + 3H2O 7H2O + Cr3+ EMBED Equation.DSMT4 Cr2O72- Balance hydrogen by adding H+ ions. 6H+ + BiO3 EMBED Equation.DSMT4 Bi3+ + 3H2O 7H2O + Cr3+ EMBED Equation.DSMT4 Cr2O72- + 14H+ Write down the number of electrons gained and lost in each half reaction 2e- + 6H+ + BiO3 EMBED Equation.DSMT4 Bi3+ + 3H2O (1) 7H2O + 2Cr3+ EMBED Equation.DSMT4 Cr2O72- + 14H+ (2) Equate the total number of electrons gained and lost by multiplying eq.(1) by 3. And then add the tow half reactions. 6e- + 8H+ + 3BiO3 EMBED Equation.DSMT4 3Bi3+ + 9H2O 7H2O + 2Cr3+ EMBED Equation.DSMT4 Cr2O72- + 14H+ 3BiO3 + 2Cr3+ EMBED Equation.DSMT4 3Bi3+ + Cr2O72- + 27

2H2O PROBLEM (Acidic Media) OCl- + S2O32- EMBED Equation.DSMT4 Cl- + S4O62- Identify the elements, which undergo oxidation and reduction and split up the reaction into oxidation and reduction half reactions. OCl- EMBED Equation.DSMT4 Cl- (reduction half reaction) S2O32- EMBED Equation.DSMT4 S4O62- (oxidation half reaction) Balance oxygen by adding H2O. OCl- EMBED Equation.DSMT4 Cl- + H2O 2 S2O32- EMBED Equation.DSMT4 S4O62- Balance hydrogen by adding H+ ions. OCl- + 2H+ EMBED Equation.DSMT4 Cl- + H2O 2 S2O32- EMBED Equation.DSMT4 S4O62- Write down the number of electrons gained and lost in each half reaction 2e- + OCl- + 2H+ EMBED Equation.DSMT4 Cl- + H2O (1) 2 S2O32- EMBED Equation.DSMT4 S4O62-+ 2e- (2) Add the two half reactions. 2e- + OCl- + 2H+ EMBED Equation.DSMT4 Cl- + H2O 2 S2O32- EMBED Equation.DSMT4 S4O62-+ 2e- 2 S2O32- + OCl- + 2H+ EMBED Equation.DSMT4 S4O62-+ CO2 PROBLEM (Acidic Media) 28

MnO4- + C2O42 EMBED Equation.DSMT4 MnO2 + CO2 Identify the elements, which undergo oxidation and reduction and split up the reaction into oxidation and reduction half reactions. MnO4- EMBED Equation.DSMT4 MnO2 (reduction half reaction) C2O42 EMBED Equation.DSMT4 CO2 (oxidation half reaction) Add two OH- ions for one each oxygen atom on appropriate side. MnO4- EMBED Equation.DSMT4 MnO2 + 4 OH- C2O42 EMBED Equation.DSMT4 CO2 Balance hydrogen by adding H2O ions. 2H2O + MnO4- EMBED Equation.DSMT4 MnO2 + 4 OH- C2O42 EMBED Equation.DSMT4 CO2 Write down the number of electron gained and lost in each half reaction 3e-+ 2H2O + MnO4- EMBED Equation.DSMT4 MnO2 + 4 OH- (1) C2O42 EMBED Equation.DSMT4 CO2 + 2e- (2) Equate the total number of electrons gained and lost by multiplying eq. (1) by 2 and eq. (2) by 3. And then add the two half reactions. 6e-+ 4H2O + 2MnO4- EMBED Equation.DSMT4 2MnO2 + 8OH- 3C2O42 EMBED Equation.DSMT4 6CO2 + 6e- 3C2O42-+2MnO4- + 4H2O EMBED Equation.DSMT4 2MnO2 + 8OH- PROBLEM (Acidic Media) MnO4- + CN- EMBED Equation.DSMT4 MnO2 + CNO- Identify the elements, which undergo oxidation and reduction and split up the reaction into oxidation and reduction half reactions. 29

MnO4- EMBED Equation.DSMT4 MnO2 (reduction half reaction) CN- EMBED Equation.DSMT4 CNO- (oxidation half reaction) Balance hydrogen and oxygen by adding H2O and OH- ions. 2H2O +MnO4- EMBED Equation.DSMT4 MnO2 +4 OH- (1) CN- +2 OH- EMBED Equation.DSMT4 CNO- + H2O (2) Write down the number of electrons gained and lost in each half reaction 3e-+ 2H2O + MnO4- EMBED Equation.DSMT4 MnO2 + 4OH- (1) CN- +2OH- EMBED Equation.DSMT4 CNO- + H2O +2e- (2) Equate the total number of electrons gained and lost by multiplying eq.(1) by 3. And then add the tow half reactions. 6e-+ 4H2O + 2MnO4- EMBED Equation.DSMT4 2MnO2 + 8OH- (1) 3CN- + 6OH- EMBED Equation.DSMT4 3CNO- + 3H2O +6e- (2) 3CN- + H2O +2MnO4- EMBED Equation.DSMT4 2MnO2+ 6OH- +3CNO- PROBLEM (Acidic Media) H3AsOs + Cr2O72- EMBED Equation.DSMT4 Cr3+ + H3AsO4 Identify the elements, which undergo oxidation and reduction and split up the reaction into oxidation and reduction half reactions. Cr2O72- EMBED Equation.DSMT4 Cr3+ (reduction half reaction) H3AsOs EMBED Equation.DSMT4 H3AsO4 (oxidation half 30

reaction) Balance oxygen by adding H2O. Cr2O72- EMBED Equation.DSMT4 Cr3+ +7 H2O H3AsOs + H2O EMBED Equation.DSMT4 H3AsO4 Balance hydrogen by adding H+ ions. 14H+ +Cr2O72- EMBED Equation.DSMT4 Cr3+ +7 H2O H3AsOs + H2O EMBED Equation.DSMT4 H3AsO4 + 2H+ Write down the number of electrons gained and lost in each half reaction. 14H+ + 6e- +Cr2O72- EMBED Equation.DSMT4 Cr3+ +7 H2O (1) H3AsOs + H2O EMBED Equation.DSMT4 H3AsO4 + 2H+ + 2e- (2) Equate the total number of electrons gained and lost by multiplying eq.(1) by 3. And then add the tow half reactions. 14H+ +6e- +Cr2O72- EMBED Equation.DSMT4 2Cr3+ +7 H2O (1) 3H3AsOs + 3H2O EMBED Equation.DSMT4 3H3AsO4 + 6H+ + 6e- (2) Cr2O72- +8H+ + 3H3AsOs EMBED Equation.DSMT4 3H3As 31