Doc Brown's Chemistry

Advanced Level Inorganic Chemistry Periodic Table Revision Notes Part 9. Group 7/17 The Halogens

9.8 More on selected halogen–halide REDOX reactions

A detailed description and explanation of redox reactions involving halogen molecules and halide ions. Oxidation and reduction are described in terms of electron loss/gain and oxidation state increase/decrease. Reactions covered include the reaction between sulfur dioxide/sulfite and halogens, the reaction between chlorate(V) and manganese(IV) oxide, the reaction of chlorine with hot and cold sodium hydroxide, the reaction between aqueous chlorine and potassium iodide solution, The reaction between iodate(V) and iodide ions in acidified aqueous solution, the oxidation of the chloride ion by acidified potassium manganate(VII) and the reduction of acidified dichromate(VI) with iodide ions.

Sub-index for this page

Oxidation state: general rules, displacement reactions, variety examples of compounds explained

Particular halogen/halide redox reactions discussed in detail

The oxidation of iodide ions by iron(III) ions

The oxidation of iron(II) to iron(III) using chlorine as the oxidising agents

The reaction between sulfur dioxide and sulfite ions and halogens

The reaction between the chlorate(V) ion and solid manganese(IV) oxide

The reaction of chlorine with hot and cold sodium hydroxide

The reaction between aqueous chlorine and potassium iodide solution

The reaction between iodate(V) and iodide ions in acidified aqueous solution

The oxidation of the chloride ion by acidified potassium manganate(VII)

The reduction of acidified dichromate(VI) with iodide ions

9.8 Oxidation and Reduction – more on redox reactions involving halogens & halide ions

See also Revision notes on Inorganic and organic REDOX reactions, oxidation state explained etc.

• Oxidation and reduction:

  • oxygen loss (reduction) and oxygen gain (oxidation) are far too limited for A level!

  • electron definitions of oxidation or reduction (electron loss or gain)

  • conceiving of 'redox' reactions as two half–reactions or half–equations, one an oxidation and the other a reduction

  • an electron acceptor is an oxidising agent (and gets reduced in its action)

  • an electron donor is a reducing agent (and gets oxidised in the process)

• Displacement reactions:

  • displacement reactions (for metals or non–metals) are often redox reactions

  • they can be 'split' into two half–reactions, note spectator ions should not be included (these are ions which do not take part in the reaction i.e. do not change their chemical 'status'!

  • the equations for them can be expressed in full formula style

    • e.g. Cl₂(aq) + KI(aq) ==> 2KCl(aq) + I_2(aq/s) 

  • OR more fundamentally as redox ionic equations with spectator ions (e.g. K⁺) excluded

    • e.g. Cl₂(aq) + 2 I^–(aq) ==> 2Cl^–(aq) + I_2(aq/s) 

    • the chlorine (molecule) is reduced, oxidation state change from 0 to –1

    • the iodide (ion) is oxidised, oxidation state change from –1 to 0

  • note on the balancing: electrons are shown in the half–cell reactions BUT not in the ionic equation, getting the right ratio of one half–cell reaction to another should ensure their transfer is 'hidden'!

• Oxidation state:

  • when dealing with atoms and simple ions, redox analysis is usually quite straight forward BUT what about when dealing with electrically neutral molecules eg H₂O and ions like PO₄^3– ?

  • these situations can redox analysed by using the concept of oxidation number or oxidation state

  • some basic rules (and learn them as quickly as possible):

    1. atoms in elements, i.e. not combined with atoms of other different elements, are considered to have an oxidation state of zero or 0

    2. in simple ions the oxidation state is the same as the charge on the ion e.g.

       • in Al³⁺ Al has an oxidation state of +3,

       • in the oxide ion O^2– has an oxidation state of –2

       • and in halide ions X^–, the oxidation state of the halogen is –1

       • Note in chemical symbols the charge sign is after the number, the charge sign is placed first for oxidation states and exam boards are strict about this sign convention)

    3. Since compounds have no overall charge, the sum of the oxidation states is zero, this is easy to follow in simple ionic compounds but not so easy at first in covalent molecules or molecular ions like PO₄^3–

    4. For covalent molecules or 'molecule ions' the most electronegative element or elements will have the negative oxidation state.

    5. irrespective of being in an ionic or covalent compound:

       • F is always –1, so O is +2 in F₂O.

    6. Cl is –1 except when combined with O or F e.g. Cl is +1 in Cl₂O

    7. for ions like ClO₄^– the charge on the ion is equal to the sum of the oxidation numbers

       • so Cl is +7, O is –2, so +7 + (4 x –2) = –1 = – for overall charge on ion

       • Note the convention:

         • for oxidation state, the sign comes before the number

         • for charge on an ion, the sign comes after the number

    8. using these rules you can work out the oxidation states of the other elements in the molecule or ion

• Oxidation states in names:

  • different compounds or ions can be formed from the same elements, so at least one of the elements is in a different oxidation state

  • so there is a need to indicate this when writing the name, but note:

    • Roman numerals are used

    • the number shows the oxidation state of the element (usually +ve)

    • the number is placed in brackets after the name, but with no space between it.

    • Examples of oxidation states for halogens in their compounds are given in a table on the introduction page.

  • Using oxidation states:

    • you can do a detailed redox analysis of any equation i.e. what is oxidised and what is reduced.

    • It helps you to balance equations when given two half–reactions

    • and note:

      • oxidation is increase in oxidation state/number (e.g. –2 to –1, –1 to 0, 0 to +1, +2 to +4)

      • reduction is decrease in oxidation state/number (e.g. –1 to –3, 0 to –1, +2 to 0, +3 to +2)

• Chlorine Cl is –1 in many simple compounds combined with a less electronegative.

  • This is related electronically to the formation of the chloride ion by chlorine gaining one electron.

    • e.g. Cl [2.8.7] + e^– ==> Cl^– [2.8.8]) e.g. in sodium chloride NaCl (Na⁺ Cl^– bond, Na is +1).

  • or sharing with one electron from a less electronegative element in covalently bonding

    • e.g. HCl (^δ+H–Cl^δ– polar bond, H is +1), PCl₅ (^δ+P–Cl^δ– polar bond, P is +5)

  • BUT when combined with the more electronegative O or F, Cl has a (+) oxidation state

  • e.g. Cl is (+1) in chlorine(I) oxide, Cl₂O (^δ+Cl–O^δ– polar bond, O is –2)

  • or (+3) in chlorine(III) fluoride, ClF₃ (^δ+Cl–F^δ– polar bond, F is –1)

    • There dozens of similar so–called 'interhalogen compounds where the most electronegative halogen carries the negative oxidation state and the positive oxidation state of the less electronegative halogen is shown with (Roman numerals) and it retains its 'elemental' name.

  • There are also a whole series of 'chlorate ions' and 'oxides', where the more electronegative oxygen results in chlorine having oxidation states of (+1), (+3), (+4), (+5) and the maximum possible (+7). The latter is the maximum ox. state for any halogen, and is dictated by using all 7 outer electrons in bonding e.g.

  • chlorate(I), ClO^–, chlorate(III), ClO₂^–, chlorate(V), ClO₃^–, chlorate(VII), ClO₄^–,

    • Note that the name of a non–metallic oxy–anion changes from ...ide to ...ate when the non–metal is combined with oxygen to form the polyatomic anion and the oxidation state is denoted after the name by a (Roman numerals number) in brackets.

  • chlorine(IV) oxide, ClO₂ (chlorine dioxide), bromine(IV) oxide, BrO₂,

  • These guidelines for chlorine also apply to bromine and iodine.

• The chlorate(V) ion, ClO₃^–
  • Oxygen is in the –2 oxidation state, so with 3 O's at (–2) each, the chlorine must be in the (+5) state to give the ion an overall charge of a single minus.

• chlorate(I), ClO^–, chlorate(VII), ClO₄^– etc. oxygen is more electronegative than chlorine.

  • (once called the hypochlorite ion and the perchlorate ion respectively)

• fluorine: F is always (–1) in all compounds,

  • because it has the highest electronegativity of any element and only one electron short of a stable noble gas structure electronically e.g. formation of fluoride ion F [2.7] + e^– ==> F^– [2.8]

  • The (–1) state relates to either gaining an electron to form the fluoride ion, F^–, e.g. in ionic sodium fluoride, NaF (Na⁺ Cl^– ionic bond, Na is +1),

  • or share with one other electron to form a covalent bond e.g. in the molecules hydrogen fluoride, HF (^δ+H–F^δ– polar bond, H is +1), or tetrafluoromethane (carbon tetrafluoride), CF₄ (^δ+C–F^δ– polar bond, C is +4).

A more detailed discussion of some halogen–halide ion REDOX reactions

See also Revision notes on Inorganic and organic REDOX reactions, oxidation state explained etc.

The oxidation of iodide ions by iron(III) ions

• Iron(III) ions oxidise iodide ions to a dark brown solution of iodine (or black solid) formed with iron(II) ions.

  • 2Fe³⁺_(aq) + 2I^–_(aq) ==> 2Fe²⁺_(aq) + I_2(aq/s)

  • This accounts for why iron(III) iodide cannot exist.

  • The oxidation state changes are ...

    • iron Fe is reduced, oxidation state changes from +3 to +2

    • and iodine I is oxidised, oxidation state changes from –1 to 0

    • The orange–brown iron(III) ion becomes the pale green iron(II) ion BUT the latter's colour is obscured by the strong dark colour of the iodine formed.

• Note that chlorine is a powerful enough oxidising agent to oxidise iron(II) ion to the iron(III) ion (see next reaction described), BUT iodine is not a strong enough oxidising agent to achieve this. It is in fact the iron(III) ion that will oxidise the iodide ion, rather than the reverse.

  • The oxidising power series for these two situations is

  • Cl₂ (E^Ø_Cl2/Cl– +1.36V) > Fe³⁺ (E^Ø_Fe3+/Fe2+ +0.77V) > I₂ (E^Ø_I2/I– +0.54V),

  • which of course is numerically paralled by the decreasing values of the standard redox potentials of the half–reactions i.e. becoming less positive as the oxidising power decreases.

The oxidation of iron(II) to iron(III) using chlorine

• The oxidation of iron(II) ions to iron(III) ions is readily achieved by chlorine – a powerful oxidising agent.

• In the reducing action of aqueous iron(II) ions, the Fe oxidation state change from +2 to +3 (oxidation) as the chlorine water readily will oxidise iron(II) to iron(III)

  • 2Fe²⁺_(aq) + Cl_2(aq) ==> 2Fe³⁺_(aq) +  2Cl^–_(aq)

  • Cl oxidation state change of 0 to –1, reduction

  • The pale green of the [Fe(H₂O)₆]²⁺_(aq) ion changes to the orange colour of the [Fe(H₂O)₆]³⁺_(aq)  ion.

  • The chlorine water itself is a very pale green, and changes to the colourless chloride ion, so the colour change associated with the oxidation state change of iron(II) to iron(III) is quite clearly seen.

  • Cross–check this reaction with the oxidation of iodide ions by iron(III) ions described in the previous redox example above.

• Note if chlorine gas is passed over hot iron(II) chloride, brown iron(III) chloride is formed.

  • 2FeCl₂(s) + Cl₂(g) ==> 2FeCl₃(s)

The reaction between sulfur dioxide and sulfite ions and halogens

• If neutral molecules or more complex ions are involved, a bit more care must be taken e.g. when the sulphur dioxide is oxidised to sulphate by bromine (or the reduction of bromine to bromide).

• SO₂(aq) + Br₂(aq) + 2H₂O(l) ===> SO₄^2–(aq) + 2Br^–(aq) + 4H⁺(aq)

• (i) the oxidation half reaction is: SO₂(aq) + 2H₂O(l) ===> SO₄^2–(aq) + 4H⁺(aq) + 2e^–

  • The sulphur changes from ox. state +4 to +6 (SO₂ to SO₄^2–), oxidation

• (ii) the reduction half–reaction is: Br₂(aq) + 2e^– ===> 2Br^–(aq)

  • Two bromine atoms (as molecule) are reduced, oxidation state changes from 0 to –1, reduction.

• The hydrogen (+1) and oxygen (–2) do not change oxidation state.

  • (i) + (ii) equals the balanced equation, 2 electrons gained and lost or an ox. state rise and fall of 2 units.

  • Bromine is the oxidising agent (gain/accept e^–s, lowered ox. state),

  • and sulphur dioxide is the reducing agent (loses e^–s, inc. ox. state of S).

• Sulphur dioxide does ionise to a small extent in water to give the sulphite ion, and adding a strong non–oxidising acid like dilute hydrochloric acid to sodium metabisulphite produces the ion, which means another equation can also adequately describe the redox change in terms of sulphur and bromine.

  • e.g. if the sulphite ion acts as the reducing agent the reaction with chlorine would be written as:

  • SO₃^2–(aq) + Cl₂(aq) + H₂O(l) ===> SO₄^2–(aq) + 2Cl^–(aq) + 2H⁺(aq)

The reaction between the chlorate(V) ion and solid manganese(IV) oxide

• If a mixture of manganese(IV) oxide, potassium hydroxide and potassium chlorate(V) is heated strongly and fuse in a crucible, the following redox reaction takes place:

• 3MnO_2(s) + 6OH^–_(aq) + ClO₃^– ==> 3MnO₄^2–_(aq) + 3H₂O_(l) + Cl^–_(aq)

• The manganate(VI) ion is formed and the ox. number changes are ...

• the oxidation of 3 x Mn from (+4) to (+6), MnO₂ ==> MnO₄^–, total 6 e's lost, oxidation.

• Is balanced by the reduction of 1 x Cl from (+5) to (–1), ClO₃^– ==> Cl^–, total 6 e's gained, reduction

• and hydrogen (+1) and oxygen (–2) do not change oxidation state.

• The chlorate(V) ion is the oxidising agent (gains/accepts e^–s, lowered ox. state of Cl) and manganese(IV) oxide is the reducing agent (loses/donates e^–s, inc. ox. state of Mn).

The reaction of chlorine with hot and cold sodium hydroxide

• In all the reactions quoted below the oxidation states of hydrogen (+1) and oxygen (–2) remain unchanged and (ii) the process descriptions are over simplified but the main reactions described provide good examples of the redox chemistry of chlorine.

  • When chlorine reacts with aqueous sodium hydroxide at least two different reaction can occur depending on the temperature and concentration of the strong base–alkali.

• With cold dilute sodium hydroxide solution alkali sodium chlorate(I) (NaClO, the bleach sodium hypochlorite) is formed as well as sodium chloride.

  • 2NaOH(aq) + Cl₂(aq) ==> NaCl(aq) + NaClO(aq) + H₂O(l)

  • 2OH^–(aq) + Cl₂(aq) ==> Cl^–(aq) + ClO^–(aq) + H₂O(l)

  • In terms of oxidation states the chlorine disproportionates from 2Cl(0) to 1Cl (–1, chloride ion) plus 1Cl(+1, chlorate(I) ion).

    • Disproportionation is the simultaneous oxidation and reduction of the same species.

  • Overall 1 electron gained, (1 ox. state unit decrease) balanced by 1 electron lost (1 ox. state unit increase).

• However, with hot concentrated sodium hydroxide solution, above 75^oC, the formation of sodium chlorate(V) predominates as well as sodium chloride.

  • 6NaOH(aq) + 3Cl₂(aq) ==> 5NaCl(aq) + NaClO₃(aq) + 3H₂O(l)

  • 6OH^–(aq) + 3Cl₂(aq) ==> 5Cl^–(aq) + ClO₃^–(aq) + 3H₂O(l)

  • In terms of oxidation state changes, the chlorine disproportionates from 6 Cl (0, element) to 5 Cl (–1, chloride ion) plus 1 Cl (+5, chlorate(V)ion).

  • Overall 5 electrons gained, (5 ox. state unit decrease) balanced by 5 electrons lost (5 ox. state unit increase).

• The change in reaction mode from cold to hot sodium hydroxide solution is due to the instability of the chlorate(I) ion, which at higher temperatures disproportionates into the chloride ion and the chlorate(V) ion.

  • 3NaClO(aq) ==> 2NaCl(aq) + NaClO₃(aq)

  • 3ClO^–(aq) ==> 2Cl^–(aq) + ClO₃^–(aq)

  • In terms of oxidation states the 'chlorine' in the chlorate ion disproportionates from 3 Cl(+1) to 2 Cl(–1, chloride ion) plus 1Cl(+5, chlorate(V) ion).

  • Overall 4 electrons gained, (4 ox. state unit decrease) balanced by 4 electrons lost (4 ox. state unit increase).

• A concentrated solution of sodium chlorate(I) is a useful source of chlorine in the laboratory because it readily reacts with conc. hydrochloric acid to give off the gas.

  • NaClO(aq) + 2HCl(aq) ==> NaCl(aq) + H₂O(l) + Cl_2(aq/g)

  • ClO^–(aq) + Cl^–(aq) + 2H⁺(aq) ==> H₂O(l) + Cl_2(aq/g)

  • In terms of oxidation states the 'chlorine' here does the opposite of disproportionation and changes from 1 Cl(+1, chlorate(I) ion) plus 1 Cl(–1, chloride ion) to give 2 Cl(0, chlorine molecule).

  • Overall 1 electron lost, (1 ox. state unit increase) balanced by  electron gained (1 ox. state unit decrease).

The reaction between aqueous chlorine and potassium iodide solution

• Half–cell reaction data:

  • (i) ¹/₂Cl₂(aq) + e^– ==> Cl^–(aq) (E^Ø = +1.36)

    • can be written as Cl₂(aq) + 2e^– ==> 2Cl^–(aq) but E^Ø still +1.36)

  • (ii) ¹/₂I₂(aq) + e^– ==> I^–(aq) (E^Ø = +0.54V)

    • can be written as I₂(aq) + 2e^– ==>  2I^–(aq) but E^Ø still +0.54V)

• In terms of oxidation state changes ...

• Chlorine molecules are reduced from ox. state (0) to (–1) of the chloride ion, 1 electron gain.

• Iodide ions are oxidised from ox. state (–1) to (0) of the iodine molecule, 1 electron loss.

• So chlorine molecules are the oxidising agent (more powerful e^– acceptor, more +ve E^Ø) and iodide ions are the reducing agent (e^– donor, less +ve E^Ø).

• 2 x oxidation half–cell, (ii) reversed	2I–(aq) ==> I2(aq) + 2e–
  2 x reduction half–cell, (i)	Cl2(aq) + 2e– ==> 2Cl–(aq)
  added gives full redox equation	Cl2(aq) +  2I–(aq) ==> Cl2(aq) + I2(aq)

• One method of estimating chlorine in water e.g. from bleaches, is to add excess potassium iodide and titrating the liberated iodine with standardised sodium thiosulphate, which itself is another redox reaction.

The reaction between iodate(V) and iodide ions in acidified aqueous solution

• Half–cell reaction data:

  • (i) ¹/₂I₂(aq) + e^– ==> I^–(aq) (E^Ø = +0.54V, iodide gets oxidised, acts as reducing agent, less positive E^Ø)

  • (ii) IO₃^–(aq) + 6H⁺(aq) + 5e^– ==> ¹/₂I_{2(aq) +} 3H₂O_(l)

    • E^Ø = +1.19V, reduction of oxidising agent, more positive E^Ø

• In terms of oxidation state changes ...

• The iodide ions (I at –1) are oxidised to iodine molecules (I at 0) by electron loss to the iodate(V) ion, I –1 to 0 ox. state.

• The iodate(V) ions (I at +5) are reduced to iodine molecules (I at 0) by electron gain from the iodide ions (the reducing agent), I +5 to 0 ox. state.

• Hydrogen (+1) and oxygen (–2) do not change oxidation state.

• 5 x ox'n half–cell, (i) reversed	5I–(aq) ==> 5/2I2(aq) + 5e–
  1 x reduction half–cell, (ii)	IO3–(aq) + 6H+(aq) + 5e– ==> 1/2I2(aq) + 3H2O(l)
  added gives full equation	IO3–(aq) + 6H+(aq) + 5I–(aq) ==> 3I2(aq) + 3H2O(l)

• The reaction can be used to estimate iodate(V) by adding excess potassium iodide and titrating the liberated iodine with standardised sodium thiosulphate or using the liberated iodine from a known quantity of potassium iodate(V) salt with excess KI(aq) salt solution to standardise the sodium thiosulphate.

The oxidation of the chloride ion by acidified potassium manganate(VII)

• Half–cell reaction data:

  • (i) MnO₄^–(aq) + 8H⁺_(aq)  + 5e^– ==> Mn²⁺(aq) + 4H₂O(l) (E^Ø = +1.52, reduction of the oxidising agent)

  • (ii) ¹/₂Cl₂(aq) + e^– ==> Cl^–(aq) (E^Ø = +1.36, lower E^Ø, so cannot act as an oxidising agent)

• Both chlorine and potassium manganate(VII) are strong oxidising agents, but chlorine is the weaker, so chloride ions are oxidised to chlorine.

• In terms of oxidation state changes ...

• Oxidation: Chlorine as the chloride ions at (–1) lose electrons to give chlorine molecules at ox. state (0).

• Reduction: Mn (+7) is reduced to Mn (+2), 5e^– gain, acts as the oxidising agent.

• Hydrogen (+1) and oxygen (–2) do not change oxidation state.

• 10 x oxi'n half–cell, (ii) rev.	10Cl–(aq) ==> 5Cl2(aq/g) + 10e–
  2 x red'n half–cell, (i)	2MnO4–(aq) + 16H+(aq)  + 10e– ==> 2Mn2+(aq) + 8H2O(l)
  added – full redox equation	2MnO4–(aq) + 16H+(aq) + 10Cl–(aq) ==> 2Mn2+(aq) + 8H2O(l) + 5Cl2(g/aq)

• This reaction is used to prepare chlorine gas (green and toxic) by running conc. hydrochloric acid on to moistened potassium manganate(VII) crystals (old name potassium permanganate).

The reduction of acidified dichromate(VI) with iodide ions

• Half–cell reaction data:

  • (i) Cr₂O₇^2–(aq) + 14H⁺(aq) + 6e^– ==> 2Cr³⁺(aq) + 7H₂O(l)

    • E^Ø = +1.33, the reduction of the oxidising agent

  • (ii) ¹/₂I₂(aq) + e^– ==> I^–(aq)

    • E^Ø = +0.54V, lower E^Ø, so cannot act as oxidising agent

• In terms of oxidation state changes ...

• Oxidation: Iodide ions at (–1) lose electron to give iodine molecules at I(0).

• Reduction: Each Cr at (+6) is reduced by gaining 3e^– to give Cr at (+3), so the Cr₂O₇^2– ion is the oxidising agent.

• Hydrogen (+1) and oxygen (–2) do not change oxidation state.

• 6 x oxidation half–cell, (ii) rev.	3I2(aq) + 6e– ==> 6I–(aq)
  1 x reduction half–cell, (i)	Cr2O72–(aq) + 14H+(aq) + 6e– ==> 2Cr3+(aq) + 7H2O(l)
  added full redox equation	Cr2O72–(aq) + 14H+(aq) + 6I–(aq) ==> 2Cr3+(aq) + 3I2(aq) + 7H2O(l)

• This reaction can be used to quantitatively measure chromium(VI) in dichromates, Cr₂O₇^2–, or chromates, CrO₄^2– (which change to dichromate(VI) on acidification, yellow ==> orange). Excess potassium iodide is added and the liberated iodine

See also Revision notes on Inorganic and organic REDOX reactions, oxidation state explained etc.

keywords phrases formula oxidation states balanced symbol equations electrode potentials: 2Fe3+(aq) + 2I–(aq) ==> 2Fe2+(aq) + I2(aq/s) Cl2 (EØCl2/Cl– +1.36V) > Fe3+ (EØFe3+/Fe2+ +0.77V) > I2 (EØI2/I– +0.54V) 2Fe2+(aq) + Cl2(aq) ==> 2Fe3+(aq) +  2Cl–(aq) 2FeCl2(s) + Cl2(g) ==> 2FeCl3(s) SO2(aq) + Br2(aq) + 2H2O(l) ==> SO42–(aq) + 2Br–(aq) + 4H+(aq) SO2(aq) + 2H2O(l) ==> SO42–(aq) + 4H+(aq) + 2e– Br2(aq) + 2e– ==> 2Br–(aq) SO32–(aq) + Cl2(aq) + H2O(l) ==> SO42–(aq) + 2Cl–(aq) + 2H+(aq) 3MnO2 + 6OH– + ClO3– ==> 3MnO42– + 3H2O + Cl– 2NaOH(aq) + Cl2(aq) ==> NaCl(aq) + NaClO(aq) + H2O(l) 2OH–(aq) + Cl2(aq) ==> Cl–(aq) + ClO–(aq) + H2O(l) 6NaOH(aq) + 3Cl2(aq) ==> 5NaCl(aq) + NaClO3(aq) + 3H2O(l) 6OH–(aq) + 3Cl2(aq) ==> 5Cl–(aq) + ClO3–(aq) + 3H2O(l) 3NaClO(aq) ==> 2NaCl(aq) + NaClO3(aq) 3ClO–(aq) ==> 2Cl–(aq) + ClO3–(aq) NaClO(aq) + 2HCl(aq) ==> NaCl(aq) + H2O(l) + Cl2(aq/g) ClO–(aq) + Cl–(aq) + 2H+(aq) ==> H2O(l) + Cl2(aq/g) 1/2Cl2(aq) + e– ==> Cl–(aq) (EØ = +1.36 Cl2(aq) + 2e– ==> 2Cl–(aq) Cl2(aq) + 2e– ==> 2Cl–(aq) 1/2I2(aq) + e– ==> I–(aq) (EØ = +0.54V) I2(aq) + 2e– ==>  2I–(aq) IO3–(aq) + 6H+(aq) + 5e– ==> 1/2I2(aq) + 3H2O(l) IO3–(aq) + 6H+(aq) + 5I–(aq) ==> 3I2(aq) + 3H2O(l) MnO4–(aq) + 8H+(aq) + 5e– ==> Mn2+(aq) + 4H2O(l) (EØ = +1.52 2MnO4–(aq) + 16H+(aq) + 10Cl–(aq) ==> 2Mn2+(aq) + 8H2O(l) + 5Cl2(g/aq) Cr2O72–(aq) + 14H+(aq) + 6I–(aq) ==> 2Cr3+(aq) + 3I2(aq) + 7H2O(l)

[SEARCH BOX] 

Website content © Dr Phil Brown 2000+. All copyrights reserved on revision notes, images, quizzes, worksheets etc. Copying of website material is NOT permitted. Exam revision summaries & references to science course specifications are unofficial.

 Doc Brown's Chemistry 

*