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TOPIC 11.
OXIDATION AND REDUCTION REACTIONS.
Early concepts of oxidation. The name "oxidation" was initially applied to reactions where substances combined with the element oxygen. Thus any substance burning in air was said to be oxidised, the product being some type of oxide. For example, burning carbon to produce carbon dioxide is an oxidation, as shown by the equation
C + O 2 v CO 2
Subsequently it was realised that reactions of substances with elements other than oxygen were of essentially the same type. For example, hydrogen can react with oxygen to form the compound water, but equally it can react with chlorine to form the compound hydrogen chloride. In both reactions the free element hydrogen is converted to a compound of hydrogen and another non-metal, and so both were classed as oxidations even though no oxygen was involved in the second case.
2H 2 + O 2 v 2H 2 O
H 2 + Cl 2 v 2HCl
The reverse reaction, conversion of compounds such as oxides of metals to the elemental metal were called "reduction" reactions, for example, the reduction of copper(II) oxide to copper by heating with charcoal (carbon).
2CuO + C v 2Cu + CO 2
The gain or loss of oxygen is still a useful way of recognising some oxidation or reduction reactions, but with a knowledge of the structure of atoms, a rather different definition is now more widely used.
Oxidation reactions as a loss of electrons. Consider the following reaction in which the metal, magnesium, is treated with hydrochloric acid, as discussed in Topic 6. The magnesium dissolves to form Mg2+ ions in solution and hydrogen gas is evolved. The equation given previously was
Mg(s) + 2H+(aq) v Mg2+(aq) + H 2 (g)
Notice that the free element magnesium (Mg) has been converted to the compound magnesium chloride, MgCl 2 , which is present in solution as its component ions Mg2+(aq) and ClS(aq). Evaporating the water would isolate this compound as shown in the following equation.
Mg2+(aq) + 2ClS(aq) v MgCl 2 (s)
This is therefore just like the examples of oxidation given above where elements were converted to compounds. However, when the reaction equation is written in this way the electronic nature of the change is more apparent. The electrically neutral Mg atoms are converted to the charged Mg2+^ cations. For this to occur, each Mg atom has lost 2 electrons according to the following HALF EQUATION :
Mg(s) v Mg2+(aq) + 2eS...... (1) Oxidation
At the same time, the H+^ ions have gained 1 electron per ion to form hydrogen atoms which have become H 2 molecules according to the following half equation.
XI - 1
OXIDATION IS THE LOSS OF ELECTRONS
REDUCTION IS THE GAIN OF ELECTRONS
2H+(aq) + 2eS^ v H 2 (g) ...... (2) Reduction
In this second half reaction the element hydrogen (H 2 ) has been formed.
It can be seen from this example that an oxidation reaction is one in which a
species loses electrons and a reduction reaction is one in which a species gains
electrons. As electrons lost by one species must be accepted by another, then both
reactions must occur simultaneously. Thus equation (1) is an OXIDATION
HALF REACTION (electrons lost by Mg) and equation (2) is a REDUCTION
HALF REACTION (electrons gained by H+). To emphasise that an oxidation is
always accompanied by a reduction, the term REDOX REACTION is used.
The overall reaction equation is obtained by combining the separate half equations
in such a way that the number of electrons lost exactly equals the number of
electrons gained, and the final equation is called a REDOX EQUATION. In this
example, the balanced overall equation results by simply adding the two half
equations.
Mg(s) + 2H+(aq) v Mg2+(aq) + H 2 (g)
Note that the 2 electrons which appeared previously on each side of the half
equations have now cancelled out.
All reactions of the type previously regarded as oxidations on the basis of gain of
oxygen atoms have the same characteristic as in the magnesium example above -
the transfer of electrons from the species oxidised to the species reduced. Thus the
most general definitions of oxidation and reduction reactions are:
The species which causes the oxidation to occur (H+^ in the above example) is
called the OXIDIZING AGENT or the OXIDANT , while the species which is
oxidized is called the REDUCING AGENT or REDUCTANT. Equations written
in the manner of (1) and (2) above are called ION-ELECTRON HALF
EQUATIONS.
The following are all examples of redox reactions.
Zn(s) + 2H+(aq) v Zn2+(aq) + H 2 (g)
2Na(s) + Br 2 (l) v 2NaBr(s)
2Fe(s) + 3Cl 2 (g) v 2FeCl 3 (s)
2HgO(s) + heat v 2Hg(l) + O 2 (g)
S(s) + O 2 (g) v SO 2 (g)
In each example, free elements were converted to compounds and/or elements
combined in compounds were converted back to the free state.
In the more complicated case of the electrolysis of water shown in Topic 1 lectures,
a solution of sulfuric acid was added to the water as there are very few H+^ and OH!
ions in pure water (see Topic 13) so little reaction would be observed. The
presence of the hydrogen ions from the acid allows conduction to occur by
electrolysis in which the H+^ ions gain electrons and are reduced to H 2 (g) at the
negative electrode. At the other electrode, sulfate ions are unable to react at the
voltages used. Instead water molecules which are attracted to the positively
charged electrode by virtue of the slight negative charge on the O atoms in the
molecule (see Topic 6), undergo the oxidation reaction. By a complex series of
steps, H 2 O molecules are converted to O atoms and then to O 2 (g) molecules. In the
process, H+^ ions are released and these exactly replace those lost at the negative
electrode. The half equations are
2H+(aq) + 2e!^ v H 2 (g) reduction
and 2H 2 O v O 2 (g) + 4H+(aq) + 4e!^ oxidation
When combined so that number of electrons gained in the reduction = the number
of electrons lost in the oxidation, the overall redox reaction equation is
4H+(aq) + 2H 2 O v 2H 2 (g) + O 2 (g) + 4H+(aq) redox
This equation shows that the same amount of H+^ appears on both sides of the
equation. While it is necessary for electrolysis to occur, there is no overall change
in the amount of H+^ as the reaction proceeds.
As electrolysis always involves transfer of electrons, then it is by definition always
a redox reaction. Electrolysis is the means by which ionic compounds conduct electricity and when reversed, is the mechanism by which is batteries operate. Note that, unlike metals, when an ionic substance conducts electricity it necessarily
undergoes chemical reactions at the electrodes and so is converted to different substances. Metals can carry an electrical current indefinitely but in the process of electrolysis the current ceases to flow once the ionic reactant has been completely
converted to its products.
In both of the above examples of electrolysis, compounds were converted to elements by means of redox reactions. Now consider the following reaction:
5Fe2+(aq) + MnO 4 S(aq) + 8H+(aq) v 5Fe3+(aq) + Mn2+(aq) + 4H 2 O(l) ...... (3)
No free element has either been formed from or converted to a compound but the Fe2+^ ion has obviously lost 1 electron to produce the Fe3+^ ion. From the above definition of oxidation, this means that Fe2+^ has been oxidised. We therefore need a method of recognising redox reactions which includes this situation where there are no free elements on either side of the equation.
The concept of oxidation number. In the above example, the half equation for the reaction involving iron is
Fe2+(aq) v Fe3+(aq) + eS
As electrons appear on the right hand side, this is clearly an oxidation. Since
aloxidations are necessarily accompanied by a reduction, then which is the species
reduced? The answer is not readily apparent because manganese is present on the
left as part of the polyatomic ion MnO 4 S^ and on the right as the simple Mn2+^ cation.
Likewise, hydrogen is present on the left as the cation H+^ and on the right as part of
the compound H 2 O. There are no uncombined elements on either side to act as a
guide. A clue is provided by noting that the Mn atom in the MnO 4 S^ ion has lost
oxygen atoms in the reaction, leaving it as the Mn2+^ ion - an indication of a
reduction reaction. However, by assigning a quantity known as the OXIDATION
NUMBER (or OXIDATION STATE ) to each atom in the equation and observing
which atoms change their oxidation numbers during the reaction, the species
oxidised and reduced can be established unequivocally. Oxidation number is the
notional charge that an atom would bear if all electrons in each bond to it were
assigned to the more electronegative atom. Oxidation numbers are not ionic
charges and, to avoid confusion, should always be written in Roman numerals with
a + or! sign as a prefix. This also serves to distinguish oxidation numbers from
Check your understanding of this section. Why is the combination of magnesium with either oxygen or chlorine regarded as the same type of reaction? Define “oxidation” and “reduction” in terms of electron transfer. Define the terms oxidant, reductant, oxidizing agent, reducing agent, ion-electron half equation. List some reduction reactions that are vital to our industrialised world. Why must an oxidation reaction be accompanied by a reduction reaction? What factor determines the stoichiometric ratio of reductant and oxidant in a balanced redox equation? Why would it not be suitable to use hydrochloric acid instead of sulfuric acid when electrolysing water to form hydrogen and oxygen gases? Could the conversion of one compound into another compound be a redox reaction? This proposition is answered in the following section.
Rules for assigning oxidation numbers.
- The oxidation number of a simple cation is the same as the cationic charge. e.g. Na+^ O.N. = +I; Mg2+^ O.N. = +II; Al3+^ O.N. = +III Use was made of this in earlier Topics to name unambiguously cations such as those of copper which could be Cu2+^ or Cu+^ and in one system of naming covalent compounds.
- The oxidation number of a simple anion is the same as the anionic charge. e.g. O^2 S^ O.N. = !II; ClS^ O.N. = !I; N^3 S^ O.N. = !III
- The oxidation number of atoms in an element = zero.
- In all compounds except peroxides and OF 2 , oxygen has the oxidation number !II, and H in most compounds has oxidation number = +I.
- The sum of all the oxidation numbers in a polyatomic ion must equal the charge on the ion, or in a compound, must add up to zero. e.g. in MnO 4 S, the sum of all oxidation numbers is (4 × !II) + Z =! 1 where Z is the oxidation number of Mn. Thus Z = +VII. or, in the molecule CO 2 , Y + (2 × !II) = 0, where Y stands for the oxidation number of carbon in the compound CO 2. Thus Y = +IV. Note that this does not imply that a C4+^ ion is present in carbon dioxide
- the carbon/oxygen bonds are in fact covalent.
Writing redox equations using ion-electron half equations.
The following procedure should be followed in writing redox equations using the
ion-electron half equation method.
- Set out the skeleton half equations for the oxidation and the reduction.
- Balance the atoms on both sides of each half equation.
- Then, balance the charge on both sides of each half equation by adding electrons as required.
- Finally, combine the oxidation and reduction half equations in the ratio which will cancel out the electrons from each side.
Example: Na v Na+^ + eS^ oxidation reaction
Cl 2 + 2eS^ v 2ClS^ reduction reaction
2Na + Cl 2 v 2Na+^ + 2ClS^ overall redox reaction
Redox reactions in biological systems. All living cells rely on redox reactions to supply the energy needed for them to remain viable and to reproduce. In the beginning on earth, there was no elemental oxygen in the atmosphere but much carbon dioxide. Thus the redox reactions utilised in respiration and photosynthesis that underlie most living systems now were not then available. Instead the first living cells used redox reactions involving elemental sulfur and hydrogen sulfide which contains the reduced form of sulfur and the oxidation of H 2 S to S provided the reduction half reaction of redox reactions that provided cellular energy. This type of cell redox reaction still operates in species that inhabit anaerobic environments such as swamps and deep sea vents. About 3.4 billion years ago the evolution of photosynthetic bacteria which use snlight to convert carbon dioxide from the atmosphere and water to carbohydrates and oxygen gas lead to the greatest atmospheric pollution to ever occur on earth and now O 2 gas constitutes about 20% of the atmosphere. As oxygen concentrations in the atmosphere rose, so the anaerobic organisms that started life on earth died out, replaced by organisms that could use the redox reactions involving elemental oxygen and water which contains its reduced form. [ Remember that the reduction of a species in a half reaction causes the oxidation of a species in the other half reaction - oxidants such as S and O 2 are necessarily reduced to H 2 S and H 2 O in that process. ] When organisms use the redox reaction in which elemental oxygen is converted to carbon dioxide as the oxidant in the cell’s redox reactions, the process is known as respiration and does not utilise light. In respiration, carbohydrates such as glucose are oxidised to carbon dioxide and water with the evolution of energy. This process necessarily occurs via many small steps because (a) releasing all the energy in one step would destroy the cell by heating and (b) the process is much more efficient when done in small steps which also produce many intermediates required for cell metabolic processes. The process of photosynthesis can be represented by the overall equation in which the O atom in water is oxidised to O 2 gas while the C atom in CO 2 is reduced
6CO 2 + 6H 2 O + light 6 C 6 H 12 O 6 + 6O 2
while respiration can be represented by the overall equation in which O 2 is reduced to CO 2
C 6 H 12 O 6 +6O 2 6 6CO 2 + 6H 2 O
where C 6 H 12 O 6 represents a carbohydrate such as glucose. However, note again that these equations are simply summarising a number of intermediate steps involving many components not shown in the equations. Respiration is in effect the reverse of photosynthesis but no light is involved.
Location of oxidants and reductants in the Periodic Table. From the definition of oxidation as a loss of electrons, it is apparent that elements which have few electrons more than a noble gas in their outer levels will be the most easily oxidised - i.e. metals. Consequently, Group 1 elements are the easiest to oxidise as they have only 1 electron more than a noble gas structure and therefore the least amount of energy is required to remove that electron. Group 2 and Group 13 elements are also mostly easy to oxidise. These elements are therefore good reducing agents, as electrons can be removed from their outer levels with only a relatively small input of energy required. In contrast, elements in Group 17 are very close to attaining the inert gas electron structure, requiring only the gain of 1
atom undergoing an oxidation will experience an increase in its oxidation number while an atom being reduced will have its oxidation number decreased. Oxidation number is not the actual charge on an atom nor is it the valency of that atom. To distinguish oxidation numbers from charge or valency, Roman numerals should be used for all oxidation numbers. Thus there are several methods of recognising a redox reaction: (i) as free elements and compounds being interconverted and especially where atoms of non-metals such as oxygen are gained or lost in that interconversion. (ii) as species gaining/losing electrons (iii) as species changing oxidation number. Elements that are the most easily oxidised (good reducing agents) are the metals from the first few families as these require relatively little energy for the removal of their outer level electrons. Conversely, the elements that are the best oxidising agents come from the families that are only a few electrons short of the noble gas electron structure, particularly elements such as oxygen and the halogens.
Recommended follow up chemcal module:
Section: General Chemistry
Module: Electrochemistry
Topic: Oxidation numbers; redox equations; galvanic cells; electrode potentials.
TUTORIAL QUESTIONS - TOPIC 11.
- Explain the following terms: oxidation half equation; reduction half equation;
oxidizing agent; oxidant; reducing agent; reductant.
- Give three definitions of "oxidation" and three definitions of "reduction".
Illustrate each with an example.
- Give the oxidation number of the underlined atom in each of the following:
(i) K+^ (ii) Al (iii) SO 42 S^ (iv) FeSO 4
(v) N 2 (vi) NaCl (vii) Na 2 SO 4 (viii) Na 3 PO 4
(ix) Na 2 CO 3 (x) MgO (xi) MgCl 2 (xii) NH 3
(xiii) NO 3 S^ (xiv) N^3 S^ (xv) CuO (xvi) Cu 2 O
(xvii) Fe 2 O 3 (xviii) Cr 2 O 72 S^ (xix) Bi3+^ (xx) CO 2
- Using the ion electron half equation method, balance the following redox equations. (i) Fe3+^ + IS^ v Fe2+^ + I 2
(ii) Sn2+^ + Fe3+^ v Sn4+^ + Fe2+
(iii) Cu + Ag+^ v Cu2+^ + Ag
(iv) H+^ + Al v H 2 + Al3+
(v) H 2 S + Fe3+^ v H+^ + S + Fe2+
- Give the oxidation number of the underlined atom in each of the following species: NiO Na 2 S MnO 4 S^ K 2 CrO 4 HNO 3
NO 2 SO 3 H 3 PO 4 H 2 ClO 4 S
- Indicate which of the following equations represent redox reactions by writing "redox" after the equation. Where the reaction is redox, give the atom oxidised, the atom reduced and their oxidation numbers.
(a) Ag+^ + ClS^ v AgCl(s)
(b) CaCO 3 (s) + 2H+^ v Ca2+^ + CO 2 (g) + H 2 O
(c) Mg(s) + 2H+^ v Mg2+^ + H 2 (g)
(d) MgO(s) + 2H+^ v Mg2+^ + H 2 O
(e) Cu(s) + S(s) v CuS(s)
- The mnemonic OILRIG is often quoted as a means of remembering what
happens to electrons in oxidation and reduction reactions. What would this mnemonic be when expanded?
- Sulfur dioxide reacts with water to form sulfite ion. Is this a redox reaction?
Justify your answer.
ANSWERS TO TUTORIAL TOPIC 11
- Oxidation half equation: An equation showing only the oxidation part of a redox reaction, balanced with respect to both atoms and charge. The charge balance is achieved by adding the required number of electrons to the right hand side of the equation.
Reduction half equation: An equation showing only the reduction part of a redox reaction, balanced with respect to both atoms and charge. The charge balance is achieved by adding the required number of electrons to the left hand side of the equation.
Oxidizing agents are any species that cause oxidation and thus are them- selves reduced - e.g. the oxidizing agent O 2 is reduced to O^2 S^ when it oxidises a metal such as Zn to Zn2+.
Oxidant: Another name for an oxidizing agent - oxidants cause an oxidation to occur and are themselves reduced in the redox reaction.
Reducing agents are any species that cause reduction and thus are them- selves oxidized - e.g. the reducing agent H 2 S is oxidized to S when it reduces Fe3+^ to Fe2+.
Reductant: Another name for a reducing agent - reductants cause a reduction to occur and are themselves oxidized in the redox reaction.
- Oxidation of a species occurs when:
(i) it gains oxygen atoms e.g. 2Mg + O 2 v 2MgO
(ii) it loses electrons e.g. Mg v Mg2+^ + 2eS
(iii) its oxidation number increases.
e.g. the oxidation number of Mg in both (i) and (ii) increases from zero on the LHS to +II on the RHS.
Reduction of a species occurs when:
(i) it loses oxygen atoms e.g. CuO + H 2 v Cu + H 2 O
(ii) it gains electrons e.g. Cu2+^ + 2eS^ v Cu
(iii) its oxidation number decreases.
e.g. the oxidation number of Cu in both (i) and (ii) decreases from +II on the LHS to zero on the RHS.
- (i) +I The O.N. of a simple cation = the charge on that cation.
(ii) 0 The O.N. of an element = zero.
(iii) +VI The sum of the O.N. of all the atoms = charge on the ion = –II.
The sum of the O.N. of the four O atoms = 4 × –II = –VIII.
à O.N. of the S atom = +VI.
(iv) +II The Fe in this compound must be Fe2+^ ion to balance the SO 42 S
ionic charge, so its O.N. = +II.
(v) 0 The O.N. of an element = zero.
(vi) +I The Na in this compound must be Na+^ to balance the ClS^ ionic
charge, so its O.N. = +I [Na, like all Group I elements, only
has the +I oxidation state.]
(vii) +I The Na in this compound must be Na+^ to balance the SO 42 S
ionic charge, so its O.N. = +I.
(viii) +V The PO 43 S^ ionic charge = sum of the O.N. of the P and the four
O atoms. The O.N. of each O atom = –II, so O.N. of the P atom
= +V.
(ix) +IV The CO 32 S^ ionic charge = sum of the O.N. of the C and the
three O atoms. The O.N. of each O atom = –II, so O.N. of the C
atom = +IV.
(x) +II The Mg in this compound must be Mg2+^ to balance the O^2 S^ ionic
charge, so its O.N. = +II.
(xi) +II The Mg in this compound must be Mg2+^ to balance the two ClS
ionic charges, so its O.N. = +II. [Mg like the other Group II
elements only has the +II oxidation state.]
(xii) !III Hydrogen normally has the +I oxidation state except in ionic
hydrides. The sum of the O.N. of the atoms = zero, so the N
atom must have O.N. = –III.
(xiii) +V The sum of the O.N. of the atoms = –I and the O atoms each
have O.N. = –II so the O.N. of the N atom = +V.
(xiv) !III The O.N. of a simple anion = charge on that anion = –III
(xv) +II The Cu in this compound must be Cu2+^ to balance the O^2 S
ionic charge, so its O.N. = +II.
(xvi) +I The Cu in this compound must be Cu+^ to balance the O^2 S^ ionic
charge, so its O.N. = +I.
(xvii) +III The Fe in this compound must be Fe3+^ to balance the total O^2 S
ionic charge, so its O.N. = +III.
(xviii) +VI The Cr 2 O 72 S^ ionic charge = sum of the O.N. of the two Cr atoms
and the seven O atoms. The O.N. of each O atom = –II, so O.N.
of each Cr atom = +VI.
HNO 3 +V The nitrate ion has the formula NO 3 S. The sum of the O.N. of
the N atom and the three O atoms = charge on the ion = –I. Given each O
atom has O.N. = –II, then the N atom has O.N. = +V.
NO 2 +IV The sum of the O.N. of the N and the two O atoms = zero. As the O.N. of each O atom = –II, then the O.N. of the N atom = +IV.
SO 3 +VI The sum of the O.N. of the S and the three O atoms = zero. As the O.N. of each O atom = –II, then the O.N. of the S atom = +VI.
H 3 PO 4 +V The phosphate ion has the formula PO 43 S. The sum of the O.N. of the P atom and the four O atoms = charge on the ion = –III. Given each O atom has O.N. = –II, then the P atom has O.N. = +V.
H 2 0 All elements have O.N. = zero.
ClO 4 S^ +VII The sum of the O.N. of the Cl atom and the four O atoms = charge on the ion = –I. Given each O atom has O.N. = –II, then the Cl atom has O.N. = +VII.
- (a) not redox (no atom changes its oxidation state).
(b) not redox (no atom changes its oxidation state).
(c) redox reaction Mg, the element (O.N. = 0) is oxidised to Mg2+^ which has O.N. = +II. The H+^ ion (O.N. = +I) is reduced to the element, H 2 , which has O.N. = 0.
(d) not redox (no atom changes its oxidation state).
(e) redox reaction. Cu is oxidised from O.N. = 0 in the element to O.N. = +II in CuS. S is reduced from O.N. = 0 in the element to O.N. = !II in CuS.
- O xidation I s L oss R eduction I s G ain (of electrons) OILRIG
- The oxidation number of S in both SO 2 and SO 32 S^ is +IV. Oxygen atoms in
both species all have O.N. = !II. As there has been no change in the oxida- tion number of any atom, this is not a redox reaction.
- The oxidation state (oxidation number) of S in these species is SO 42 S^ (+VI) SO 32 S^ (+IV) S(s) (0) SO 2 (+IV) H 2 S (!II) (derived using the O.N. of O atoms = –II and H atoms = +I). Thus S has its highest oxidation number in SO 42 S^ and its lowest in H 2 S.
- The oxidation state (oxidation number) of N in these species is NO 2 (+IV) NO (+II) HNO 3 (+V) N 2 (0)
HNO 2 (+III) N 2 O (+I) Na 3 N (!III) Thus N has its highest oxidation number in HNO 3 and its lowest in Na 3 N.
- (a) Reducing agent: ClS^ Oxidizing agent: MnO 2
Atom oxidized: Cl as ClS^ (O.N. = !I) to Cl in Cl 2 (O.N. = 0). Atom reduced: Mn in MnO 2 (O.N. = +IV) to Mn as Mn2+^ (O.N. = +II).
(b) Reducing agent: SO 2 Oxidizing agent: NO 3 S Atom oxidized: S in SO 2 (O.N. = +IV) to S in SO 42 S^ (O.N. = +VI). Atom reduced: N in NO 3 S^ (O.N. = +V) to N in NO (O.N. = +II).
(c) Reducing agent: Ag Oxidizing agent: H 2 S Atom oxidized: Ag as the element (O.N. = 0) to Ag+^ in Ag 2 S (O.N. = +I). Atom reduced: H in H 2 S (O.N. = +I) to H in H 2 (O.N. = 0).
(d) Reducing agent: H 2 S Oxidizing agent: NO 3 S Atom oxidized: S in H 2 S (O.N. = !II) to S as the element (O.N. = 0). Atom reduced: N in NO 3 S^ (O.N. = +V) to N in NO 2 (O.N. = +IV).
(e) Reducing agent: Fe2+^ Oxidizing agent: MnO 4 S Atom oxidized: Fe as Fe2+^ (O.N. = +II) to Fe as Fe3+^ (O.N. = +III). Atom reduced: Mn in MnO 4 S^ (O.N. = +VII) to Mn as Mn2+^ (O.N. = +II).
(f) Reducing agent: Fe2+^ Oxidizing agent: Cr 2 O 72 S Atom oxidized: Fe as Fe2+^ (O.N. = +II) to Fe as Fe3+^ (O.N. = +III). Atom reduced: Cr in Cr 2 O 72 S^ (O.N. = +VI) to Cr as Cr3+^ (O.N. = +III).
(g) Reducing agent: BrS^ Oxidizing agent: Cl 2 Atom oxidized: Br as BrS^ (O.N. = !I) to Br as the element Br 2 (O.N. = 0). Atom reduced: Cl as the element Cl 2 (O.N. = 0) to Cl as ClS^ (O.N. = !I).
(h) Reducing agent: H 2 Oxidizing agent: CuO Atom oxidized: H as the element H 2 (O.N. = 0) to H in H 2 O (O.N. = +I). Atom reduced: Cu in CuO (O.N. = +II) to Cu as the element (O.N. = 0).
(i) Reducing agent: Sn2+^ Oxidizing agent: H 3 AsO 4 Atom oxidized: Sn as Sn2+^ (O.N. = +II) to Sn as Sn4+^ (O.N. = +IV). Atom reduced: As in H 3 AsO 4 (O.N. = +V) to As in H 3 AsO 3 (O.N. = +III).
(j) Reducing agent: Pb Oxidizing agent: PbO 2 Atom oxidized: Pb as the element (O.N. = 0) to Pb as Pb2+^ (O.N. = +II). Atom reduced: Pb in PbO 2 (O.N. = +IV) to Pb2+^ (O.N. = +II). (This example shows that two different oxidation states of the same element can lead to both oxidation and reduction of the one type of atom; this reaction is the basis for the lead acid battery).
