1 / 96

Oxidation numbers and balancing equations

Oxidation was originally named for dealing with processes where oxygen was transferred during reactions. Oxidation was originally named for dealing with processes where oxygen was transferred during reactions, as in H2 CuO ? Cu H2O. Oxidation was originally named for dealing with proc

lawrencia
Télécharger la présentation

Oxidation numbers and balancing equations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

    1. Oxidation numbers and balancing equations

    2. Oxidation was originally named for dealing with processes where oxygen was transferred during reactions

    3. Oxidation was originally named for dealing with processes where oxygen was transferred during reactions, as in H2 + CuO ? Cu + H2O

    4. Oxidation was originally named for dealing with processes where oxygen was transferred during reactions, as in H2 + CuO ? Cu + H2O Oxidation is a process, which always occurs simultaneously with Reduction, and which characterises many chemical reactions.

    5. As chemistry developed, it was found that many reactions, not necessarily involving oxygen, shared very similar characteristics.

    6. As chemistry developed, it was found that many reactions, not necessarily involving oxygen, shared very similar characteristics. The words oxidation and reduction have remained.

    7. As chemistry developed, it was found that many reactions, not necessarily involving oxygen, shared very similar characteristics. The words oxidation and reduction have remained. Reactions of this type are abbreviated to REDOX reactions. They all involve a transfer of electrons from one particle to another.

    8. As chemistry developed, it was found that many reactions, not necessarily involving oxygen, shared very similar characteristics. The words oxidation and reduction have remained. Reactions of this type are abbreviated to REDOX reactions. They all involve a transfer of electrons from one particle to another. We have developed the idea of oxidation number to help in dealing with these reactions.

    9. It is easiest to start by looking at ionic reactions where there is an obvious transfer of electrons from one atom to another.

    10. It is easiest to start by looking at ionic reactions where there is an obvious transfer of electrons from one atom to another. E.g. The reaction in which ordinary salt, sodium chloride, is formed from the elements sodium and chlorine.

    11. It is easiest to start by looking at ionic reactions where there is an obvious transfer of electrons from one atom to another. E.g. The reaction in which ordinary salt, sodium chloride, is formed from the elements sodium and chlorine. We write the reaction formally as 2Na + Cl2 = 2NaCl

    12. It is easiest to start by looking at ionic reactions where there is an obvious transfer of electrons from one atom to another. E.g. The reaction in which ordinary salt, sodium chloride, is formed from the elements sodium and chlorine. We write the reaction formally as 2Na + Cl2 = 2NaCl but we also realise that sodium chloride contains only sodium and chloride ions, there are no separate sodium chloride molecules.

    13. We can then write ?2e-? 2Na + Cl2 = 2Na++ 2Cl- atoms molecules ions ions

    14. We can then write ?2e-? 2Na + Cl2 = 2Na++ 2Cl- atoms molecules ions ions where the sodium atoms have lost electrons to become Na+ ions and these electrons have been transferred to the chlorine atoms, making them into Cl- ions.

    15. We can then write ?2e-? 2Na + Cl2 = 2Na++ 2Cl- atoms molecules ions ions where the sodium atoms have lost electrons to become Na+ ions and these electrons have been transferred to the chlorine atoms, making them into Cl- ions. In the reaction the sodium atoms had an increase of one unit in their positive charge (from 0 to +1) and the chlorine a one unit decrease (from 0 to -1). Overall change is +2 for the two sodiums and -2 for the two chlorines = 0 total

    16. For ionic reactions like this one, we associate the loss of electrons as oxidation and the corresponding gain of electrons as reduction.

    17. For ionic reactions like this one, we associate the loss of electrons as oxidation and the corresponding gain of electrons as reduction. There is a useful mnemonic for remembering this idea, namely OILRIG. Which comes from the initial letters of the words in the sentence:- Oxidation is the loss of electrons, reduction is the gain of electrons.

    18. We choose this definition because of the similarity between, for example, the sodium in its reaction with chlorine and its behaviour in the reaction:-

    19. We choose this definition because of the similarity between, for example, the sodium in its reaction with chlorine and its behaviour in the reaction:- ?4e-? 4Na + O2 ? 2Na2O ( 4Na+ + 2O2- )

    20. We choose this definition because of the similarity between, for example, the sodium in its reaction with chlorine and its behaviour in the reaction:- ?4e-? 4Na + O2 ? 2Na2O ( 4Na+ + 2O2- ) where the sodium oxide is also an ionic compound like sodium chloride.

    21. We choose this definition because of the similarity between, for example, the sodium in its reaction with chlorine and its behaviour in the reaction:- ?4e-? 4Na + O2 ? 2Na2O ( 4Na+ + 2O2- ) where the sodium oxide is also an ionic compound like sodium chloride. Or, our original example ? 2e- ? H2 + CuO + ? Cu + H2O where copper ions become copper metal

    22. We define Oxidation Number (ON) in this way:

    23. We define Oxidation Number (ON) in this way: Equal to the ionic charge for simple ions

    24. We define Oxidation Number (ON) in this way: Equal to the ionic charge for simple ions e.g. Fe3+ has ON +3 and S2- has ON -2

    25. We define Oxidation Number (ON) in this way: Equal to the ionic charge for simple ions e.g. Fe3+ has ON +3 and S2- has ON -2 In most compounds H has ON +1 and O has ON -2.

    26. We define Oxidation Number (ON) in this way: Equal to the ionic charge for simple ions e.g. Fe3+ has ON +3 and S2- has ON -2 In most compounds H has ON +1 and O has ON -2. (Exceptions are metal hydrides where ON H = -1 and peroxides where ON O = -1)

    27. We define Oxidation Number (ON) in this way: Equal to the ionic charge for simple ions e.g. Fe3+ has ON +3 and S2- has ON -2 In most compounds H has ON +1 and O has ON -2. (Exceptions are metal hydrides where ON H = -1 and peroxides where ON O = -1) The total ON for a compound is always zero.

    28. We define Oxidation Number (ON) in this way: Equal to the ionic charge for simple ions e.g. Fe3+ has ON +3 and S2- has ON -2 In most compounds H has ON +1 and O has ON -2. (Exceptions are metal hydrides where ON H = -1 and peroxides where ON O = -1) The total ON for a compound is always zero. 4 The ON of an unreacted element is zero.

    29. We define Oxidation Number (ON) in this way: Equal to the ionic charge for simple ions e.g. Fe3+ has ON +3 and S2- has ON -2 In most compounds H has ON +1 and O has ON -2. (Exceptions are metal hydrides where ON H = -1 and peroxides where ON O = -1) The total ON for a compound is always zero. 4 The ON of an unreacted element is zero. 5 The overall ON change in a reaction is zero.

    30. We define Oxidation Number (ON) in this way: Equal to the ionic charge for simple ions e.g. Fe3+ has ON +3 and S2- has ON -2 In most compounds H has ON +1 and O has ON -2. (Exceptions are metal hydrides where ON H = -1 and peroxides where ON O = -1) The total ON for a compound is always zero. 4 The ON of an unreacted element is zero. The overall ON change in a reaction is zero. Some atoms with variable ON have the value shown by a Roman numeral, e.g. FeIIICl3

    31. So why is this idea useful in constructing equations?

    32. So why is this idea useful in constructing equations? Since the overall ON change is zero, this fixes the ratio of the numbers of the reduced atoms to those of the oxidised atoms.

    33. So why is this idea useful in constructing equations? Since the overall ON change is zero, this fixes the ratio of the numbers of the reduced atoms to those of the oxidised atoms. In our first example, there had to be equal numbers of sodium and chlorine atoms involved since the ON of Na went up by 1 unit (0?+1) and that of Cl went down by 1 unit (0?-1). 2Na + Cl2 = 2NaCl

    34. Here is another ionic reaction FeIICl2 + Cl2 = FeIIICl3

    35. Here is another ionic reaction FeIICl2 + Cl2 = FeIIICl3 ON +2 -1 0 +3 -1

    36. Here is another ionic reaction FeIICl2 + Cl2 = FeIIICl3 ON +2 -1 0 +3 -1 Each iron atom increases its ON by 1 (+2?+3), whilst the elemental chlorine atoms decrease by 1 (0?-1).

    37. Here is another ionic reaction FeIICl2 + Cl2 = FeIIICl3 ON +2 -1 0 +3 -1 Each iron atom increases its ON by 1 (+2?+3), whilst the elemental chlorine atoms decrease by 1 (0?-1). For an ON change of zero, this means that there must be equal numbers of FeIICl2 units and reacting Cl atoms. Since these atoms come in pairs (Cl2) the balanced equation MUST start with 2 in front of the FeCl2:-

    38. 2FeIICl2 + 1Cl2 = ...FeIIICl3

    39. 2FeIICl2 + 1Cl2 = ...FeIIICl3 and this means that the balanced equation has to be 2FeIICl2 + Cl2 = 2FeIIICl3

    40. 2FeIICl2 + 1Cl2 = ...FeIIICl3 and this means that the balanced equation has to be 2FeIICl2 + Cl2 = 2FeIIICl3 This was a long-winded solution to an easy problem, but it shows how the method works.

    41. 2FeIICl2 + 1Cl2 = ...FeIIICl3 and this means that the balanced equation has to be 2FeIICl2 + Cl2 = 2FeIIICl3 This was a long-winded solution to an easy problem, but it shows how the method works. It is more useful in balancing this reactions equation:- Copper + nitric acid ? Copper nitrate + nitric oxide + water

    42. 2FeIICl2 + 1Cl2 = ...FeIIICl3 and this means that the balanced equation has to be 2FeIICl2 + Cl2 = 2FeIIICl3 This was a long-winded solution to an easy problem, but it shows how the method works. It is more useful in balancing this reactions equation:- Copper + nitric acid ? Copper nitrate + nitric oxide + water Cu + HNO3 ? Cu(NO3)2 + NO + H2O

    43. Cu + HNO3 ? Cu(NO3)2 + NO + H2O ON 0 +1+5-2 +2 +5-2 +2-2 +1 -2

    44. Cu + HNO3 ? Cu(NO3)2 + NO + H2O ON 0 +1+5-2 +2 +5-2 +2-2 +1 -2 For HNO3, the ON of nitrogen is +5 since the H is +1 and the O is -2. Total ON = 0 = +1 + 3(-2) + ON of nitrogen 0 = -5 + ON of nitrogen ON of nitrogen = +5

    45. Cu + HNO3 ? Cu(NO3)2 + NO + H2O ON 0 +1+5-2 +2 +5-2 +2-2 +1 -2 For HNO3, the ON of nitrogen is +5 since the H is +1 and the O is -2. Total ON = 0 = +1 + 3(-2) + ON of nitrogen 0 = -5 + ON of nitrogen ON of nitrogen = +5 We can see that the oxidation number of nitrogen changes from +5?+2 (3 steps down) and that of copper from 0?+2(two steps up).

    46. This means that for every three copper atoms oxidised, two nitrogen atoms are reduced and we can start the balancing:- 3Cu + ...HNO3 ? 3Cu(NO3)2 + 2NO + ...H2O

    47. This means that for every three copper atoms oxidised, two nitrogen atoms are reduced and we can start the balancing:- 3Cu + ...HNO3 ? 3Cu(NO3)2 + 2NO + ...H2O The next step might be to check the nitrate groups, six on the right hand side: giving eight nitrogens all together, coming from nitric acid. 3Cu + 8HNO3 ? 3Cu(NO3)2 + 2NO + ...H2O

    48. This means that for every three copper atoms oxidised, two nitrogen atoms are reduced and we can start the balancing:- 3Cu + ...HNO3 ? 3Cu(NO3)2 + 2NO + ...H2O The next step might be to check the nitrate groups, six on the right hand side: giving eight nitrogens all together, coming from nitric acid. 3Cu + 8HNO3 ? 3Cu(NO3)2 + 2NO + ...H2O Now we just need to organise the water formed from the eight hydrogens 3Cu + 8HNO3 ? 3Cu(NO3)2 + 2NO + 4H2O

    49. Although this took some time, we knew that the Cu/NO ratio was fixed at 3/2 and this left only a few options for the remainder.

    50. Although this took some time, we knew that the Cu/NO ratio was fixed at 3/2 and this left only a few options for the remainder. We can apply this method to a very similar reaction:- Cu + HNO3 ? Cu(NO3)2 + NO2 + H2O

    51. Although this took some time, we knew that the Cu/NO ratio was fixed at 3/2 and this left only a few options for the remainder. We can apply this method to a very similar reaction:- Cu + HNO3 ? Cu(NO3)2 + NO2 + H2O The only difference being in the different nitrogen oxide produced.

    52. Although this took some time, we knew that the Cu/NO ratio was fixed at 3/2 and this left only a few options for the remainder. We can apply this method to a very similar reaction:- Cu + HNO3 ? Cu(NO3)2 + NO2 + H2O The only difference being in the different nitrogen oxide produced. Cu + HNO3 ? Cu(NO3)2 + NO2 + H2O ON 0 +1+5-2 +2 +5-2 +4-2 +1 -2

    53. Now the nitrogen changes only by one unit, from +5 to +4. This fixes the Cu/ NO2 ratio as 1Cu / 2NO2 So now we can start the balancing Cu + ...HNO3 ? Cu(NO3)2 + 2NO2 + ...H2O

    54. Now the nitrogen changes only by one unit, from +5 to +4. This fixes the Cu/ NO2 ratio as 1Cu / 2NO2 So now we can start the balancing Cu + ...HNO3 ? Cu(NO3)2 + 2NO2 + ...H2O There are now 4 nitrogens on the right and so Cu + 4HNO3 ? Cu(NO3)2 + 2NO2 + ...H2O

    55. Now the nitrogen changes only by one unit, from +5 to +4. This fixes the Cu/ NO2 ratio as 1Cu / 2NO2 So now we can start the balancing Cu + ...HNO3 ? Cu(NO3)2 + 2NO2 + ...H2O There are now 4 nitrogens on the right and so Cu + 4HNO3 ? Cu(NO3)2 + 2NO2 + ...H2O and finally Cu + 4HNO3 ? Cu(NO3)2 + 2NO2 + 2H2O

    56. We could have written this equation in its ionic form

    57. We could have written this equation in its ionic form as:- Cu + 2NO3- + 4H+ ? Cu2+ + 2NO2 + H2O

    58. We could have written this equation in its ionic form as:- Cu + 2NO3- + 4H+ ? Cu2+ + 2NO2 + H2O which makes it clear that the Cu/N ratio is 1/2

    59. We could have written this equation in its ionic form as:- Cu + 2NO3- + 4H+ ? Cu2+ + 2NO2 + H2O which makes it clear that the Cu/N ratio is 1/2 remember that the charges have to balance, as well as the numbers of atoms, in an ionic reaction. Here we have a net +2 charge on each side.

    60. As a last example we can look at a reaction where more than one atom is oxidised As2S3 + Cl2 ? AsCl5 + S8

    61. As a last example we can look at a reaction where more than one atom is oxidised As2S3 + Cl2 ? AsCl5 + S8 In this reaction there are two complications. Firstly, both arsenic and sulphur atoms are oxidised by chlorine

    62. As a last example we can look at a reaction where more than one atom is oxidised As2S3 + Cl2 ? AsCl5 + S8 In this reaction there are two complications. Firstly, both arsenic and sulphur atoms are oxidised by chlorine and, secondly, the sulphur atoms in elemental sulphur come in rings of eight.

    63. As a last example we can look at a reaction where more than one atom is oxidised As2S3 + Cl2 ? AsCl5 + S8 In this reaction there are two complications. Firstly, both arsenic and sulphur atoms are oxidised by chlorine and, secondly, the sulphur atoms in elemental sulphur come in rings of eight. It is easiest to balance the reaction using single sulphur atoms and then multiply everything by eight.

    64. As2S3 + Cl2 ? AsCl5 + S ON +3 -2 0 +5 -1 0

    65. As2S3 + Cl2 ? AsCl5 + S ON +3 -2 0 +5 -1 0 The overall oxidation number increase for As is +2 for the each of the two atoms (= +4) and for sulphur it is +2 for each of the three atoms (= +6).

    66. As2S3 + Cl2 ? AsCl5 + S ON +3 -2 0 +5 -1 0 The overall oxidation number increase for As is +2 for the each of the two atoms (= +4) and for sulphur it is +2 for each of the three atoms (= +6). I.e. each As2S3 molecule requires 10 (6+4) ON steps. Each chlorine atom decreases its ON by one unit, so this will require five chlorine molecules, each with two atoms

    67. As2S3 + Cl2 ? AsCl5 + S ON +3 -2 0 +5 -1 0 The overall oxidation number increase for As is +2 for the each of the two atoms (= +4) and for sulphur it is +2 for each of the three atoms (= +6). I.e. each As2S3 molecule requires 10 (6+4) ON steps. Each chlorine atom decreases its ON by one unit, so this will require five chlorine molecules, each with two atoms. This fixes the As2S3/Cl2 ratio at 1:5

    68. As2S3 + 5Cl2 ? ...AsCl5 + ...S

    69. As2S3 + 5Cl2 ? ...AsCl5 + ...S Now we can balance the rest of the atoms As As2S3 + 5Cl2 ? 2AsCl5 + ...S

    70. As2S3 + 5Cl2 ? ...AsCl5 + ...S Now we can balance the rest of the atoms As As2S3 + 5Cl2 ? 2AsCl5 + ...S S As2S3 + 5Cl2 ? 2AsCl5 + 3S

    71. As2S3 + 5Cl2 ? ...AsCl5 + ...S Now we can balance the rest of the atoms As As2S3 + 5Cl2 ? 2AsCl5 + ...S S As2S3 + 5Cl2 ? 2AsCl5 + 3S To complete the process, we have to allow the sulphurs to be in groups of eight and so we have to multiply each entry by eight:- 8As2S3 + 40Cl2 ? 16AsCl5 + 3S8

    72. Ionic reactions in solution sometimes need you to realise that water molecules, H+ ions and/or OH- ions may be present.

    73. Ionic reactions in solution sometimes need you to realise that water molecules, H+ ions and/or OH- ions may be present. Potassium permanganate, KMnVIIO4 contains MnVIIO4- ions. These are oxidising agents in acidic, neutral or basic solutions and give different products in each.

    74. Ionic reactions in solution sometimes need you to realise that water molecules, H+ ions and/or OH- ions may be present. Potassium permanganate, KMnVIIO4 contains MnVIIO4- ions. These are oxidising agents in acidic, neutral or basic solutions and give different products in each. The ions have their strongest oxidising power in acidic solution, where they are reduced to Mn2+ ions, giving an ON change of -5 (from +7 to +2).

    75. Lets see how these ions oxidise iron(II), Fe2+,ions to iron(III), Fe3+, ions.

    76. Lets see how these ions oxidise iron(II), Fe2+,ions to iron(III), Fe3+, ions. This is a single ON step and so we will need only one MnO4- ion to deal with 5 Fe2+ions.

    77. Lets see how these ions oxidise iron(II), Fe2+,ions to iron(III), Fe3+, ions. This is a single ON step and so we will need only one MnO4- ion to deal with 5 Fe2+ions. 5Fe2+ + MnO4- ? 5Fe3+ + Mn2+

    78. Lets see how these ions oxidise iron(II), Fe2+,ions to iron(III), Fe3+, ions. This is a single ON step and so we will need only one MnO4- ion to deal with 5 Fe2+ions. 5Fe2+ + MnO4- ? 5Fe3+ + Mn2+ The question now is, how to deal with the oxygens?

    79. Lets see how these ions oxidise iron(II), Fe2+,ions to iron(III), Fe3+, ions. This is a single ON step and so we will need only one MnO4- ion to deal with 5 Fe2+ions. 5Fe2+ + MnO4- ? 5Fe3+ + Mn2+ The question now is, how to deal with the oxygens? Since the reaction is in acidic solution, there will be H+ ions present and these will react to form water with the oxygen.

    80. 5Fe2+ + MnO4- + ....H+ ? 5Fe3+ + Mn2+ + ...H2O

    81. 5Fe2+ + MnO4- + ....H+ ? 5Fe3+ + Mn2+ + ...H2O The four oxygens will produce four water molecules, demanding eight hydrogens as the final step:-

    82. 5Fe2+ + MnO4- + ....H+ ? 5Fe3+ + Mn2+ + ...H2O The four oxygens will produce four water molecules, demanding eight hydrogens as the final step:- 5Fe2+ + MnO4- + ....H+ ? 5Fe3+ + Mn2+ + 4H2O

    83. 5Fe2+ + MnO4- + ....H+ ? 5Fe3+ + Mn2+ + ...H2O The four oxygens will produce four water molecules, demanding eight hydrogens as the final step:- 5Fe2+ + MnO4- + ....H+ ? 5Fe3+ + Mn2+ + 4H2O 5Fe2+ + MnO4- + 8H+ ? 5Fe3+ + Mn2+ + 4H2O

    84. If we wanted a full equation, rather than the ionic one, we need to specify both the iron compound and the acid. Lets choose ironII sulphate and sulphuric acid. The equation then becomes 10FeSO4 + 2KMnO4 + 8H2SO4 ? 5Fe2(SO4)3 + 2MnSO4 + 8H2O Where the ratio Fe/Mn is still 1:5 as before.

    85. In neutral solution the manganese becomes MnIVO2 and so there are three reduction steps from MnO4- and we can start writing the equation as

    86. In neutral solution the manganese becomes MnIVO2 and so there are three reduction steps from MnO4- and we can start writing the equation as 3Fe2+ + MnO4- + ....H+ ? 3Fe3+ + MnO2 + ...H2O

    87. In neutral solution the manganese becomes MnIVO2 and so there are three reduction steps from MnO4- and we can start writing the equation as 3Fe2+ + MnO4- + ....H+ ? 3Fe3+ + MnO2 + ...H2O Now we just need to use the hydrogen ions to balance as before. The two unbalanced oxygens need four H+, always present in water.

    88. In neutral solution the manganese becomes MnIVO2 and so there are three reduction steps from MnO4- and we can start writing the equation as 3Fe2+ + MnO4- + ....H+ ? 3Fe3+ + MnO2 + ...H2O Now we just need to use the hydrogen ions to balance as before. The two unbalanced oxygens need four H+, always present in water. 3Fe2+ + MnO4- + 4H+ ? 3Fe3+ + MnO2 + 2H2O

    89. In neutral solution the manganese becomes MnIVO2 and so there are three reduction steps from MnO4- and we can start writing the equation as 3Fe2+ + MnO4- + ....H+ ? 3Fe3+ + MnO2 + ...H2O Now we just need to use the hydrogen ions to balance as before. The two unbalanced oxygens need four H+, always present in water. 3Fe2+ + MnO4- + 4H+ ? 3Fe3+ + MnO2 + 2H2O And we can notice that this has also balanced the charges

    90. The third condition, basic solution, reduces the manganese to MnVIO42-, with an ON change of just -1 for the manganese this time.

    91. The third condition, basic solution, reduces the manganese to MnVIO42-, with an ON change of just -1 for the manganese this time. So with sulphite ions being oxidised to sulphate, with all the changing oxidation numbers shown in Roman numerals ...MnVIIO4- + SIVO32- ? ...MnVI O42- + SVIO42-

    92. The third condition, basic solution, reduces the manganese to MnVIO42-, with an ON change of just -1 for the manganese this time. So with sulphite ions being oxidised to sulphate, with all the changing oxidation numbers shown in Roman numerals ...MnVIIO4- + SIVO32- ? ...MnVI O42- + SVIO42- 2MnVIIO4- + SIVO32- ? 2MnVI O42- + SVIO42-

    93. 2MnVIIO4- + SIVO32- ? 2MnVI O42- + SVIO42- Now we need to make use of the fact that this is in basic solution and use hydroxide ions to provide the extra oxygen, with the hydrogen ending up as water.

    94. 2MnVIIO4- + SIVO32- ? 2MnVI O42- + SVIO42- Now we need to make use of the fact that this is in basic solution and use hydroxide ions to provide the extra oxygen, with the hydrogen ending up as water. 2MnVIIO4- + SIVO32- + 2OH- ? 2MnVIO42- + SVIO42- + H2O

    95. 2MnVIIO4- + SIVO32- ? 2MnVI O42- + SVIO42- Now we need to make use of the fact that this is in basic solution and use hydroxide ions to provide the extra oxygen, with the hydrogen ending up as water. 2MnVIIO4- + SIVO32- + 2OH- ? 2MnVIO42- + SVIO42- + H2O If these had all been potassium salts, a full equation would be:- 2KMnO4 + K2SO3 + 2KOH ? 2K2MnO4 + K2SO4 + H2O

    96. There are practice equations to balance on the associated pdf file. Good luck!

More Related