3A Redox Reaction

7/30/2019 3A Redox Reaction

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CHAPTER 3

Oxidation

and

Reduction


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(A) Redox Reactions, RRAfter this lesson, you should be able to:

State what oxidation is State what reduction is

Explain what redox reaction is

State what oxidizing agent is

State what reducing agent is

Calculate the oxidation number of an element in acompound

Relate the oxidation number of an element to the name of itscompound using the IUPAC nomenclature

Explain with examples oxidation and reduction processes interms of the change in oxidation number

Explain with examples oxidation and reduction processes interms of electron transfer

Explain with examples oxidizing agents and reducing agentsin redox reactions

Write oxidation and reduction half-equations and ionicequations.


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What is Redox Reaction?

Redox reaction are chemical

reactions involving oxidationand reduction occurring

simultaneously.


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Explanation of

Redox Reaction, RR

Redox reactions can be explained in term of:

Loss or gain of oxygen

Loss or gain of hydrogen

Transfer of electrons

Changes in oxidation number


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Explanation of RR based on

Loss or Gain of Oxygen

Oxidation is a chemical reaction in which oxygenis added to a substance

Reduction is defined as the loss of oxygen from asubstance

The substance that causes oxidation is called

oxidizing agent (oxidant)

The substance that causes reduction is called thereducing agent (reductant)


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2CuO(s) + C(s) 2Cu(s) + CO2(g)

CuO is reduced to Cu

C is oxidized to CO2

CuO acts as oxidizing agent (oxidant)

C acts as reducing agent (reductant)

Gains oxygen

(oxidation)

Loses oxygen

(reduction)

Example 1:


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Loss or Gain of Hydrogen

Oxidation is the loss of hydrogen from a

substance

Reduction is the gain of hydrogen from a

substance


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H2S(g) + Cl2 (g) S(s) + 2HCl(g)

H2S is oxidized to S

Cl2 is reduced to HCl

Cl2 acts as oxidizing agent (oxidant)

H2S acts as reducing agent (reductant)

Gains hydrogen

(reduction)

Loses hydrogen

(oxidation)

Example 2:


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Practice A1:

Study the following equations and identify theoxidized substances, reduced substances, oxidant

and reductant.

a)2HBr(aq) + Cl2(l) 2HCl(aq) + Br2(l)

b)Mg(s) + PbO(s) MgO(s) + Pb(s)

c)CH4(g) + Cl2(g)

CH3Cl(g) + HCl(g)d)Fe3O4(s) + 4CO(g) 3Fe(s) + 4CO2(g)

e)PbS(s) + 4H2O2(aq) PbSO4(s) + 4H2O(l)


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Transfer of Electrons

Oxidation is the loss of electrons

Reduction is the gain of electrons

Oxidizing Agentis electron acceptors

Reducing Agentis electron donors


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Zn(s) + Cu2+ (s) Zn2+ (aq) + Cu(s)

Loss e- Zn(s)

Zn

2+

(aq) + 2e-

Gain e- Cu2+ (aq) + 2e- Cu(s)

-------------------------------------------------------------

Zn(s) + Cu2+

(s) + 2e-

Zn2+

(aq) + Cu(s) + 2e-----------------------------------------------------------------------

Ionic Eq. Zn(s) + Cu2+ (s) Zn2+ (aq) + Cu(s)

=========================================

Gains electron

(reduction)

Loses electron

(oxidation)

Example 3:


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Practice A2:Study the following redox reactions

a) Cu(s) + 2Ag+(aq) Cu2+ (aq) + 2Ag(s)b) Cl2(g) + 2Br

(aq) 2Cl(aq) + Br2(l)

c) Ca(s) + 2HCl(aq) CaCl2(aq) + H2(g)

d) 2Na(s) + Cl2(g) 2NaCl(s)

e) 2Fe2+(aq) + Br2(aq) 2Fe3+(aq) + 2Br(aq)

For each of the above reaction above,

i) Write the half equations

ii) Identify the Oxidized substance

Reduced substance

Oxidizing agent

Reducing agent based on the transfer of electrons


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What is an Oxidation Number?

The oxidation number or oxidation

state of an element is the charge that

the atom of the element would have if

complete transfer of electron occurs.


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Rules in assigning Oxidation Number

Rule 1:

The oxidation number of an atom in its

elemental state is zero.

For example:

The oxidation number of each atom in

Mg, Cu, Na, H2, O2, Cl2 and P4 is zero.


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Rules in assigning Oxidation

NumberRule 2:

The oxidation number of monoatomic ion is equal

to its chargeFor example:

Ion Na+ Mg2+ Al3+ Br S2 N3

Oxidation

Number+1 +2 +3 1 2 3


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Rule 3:

The oxidation number of hydrogen in a compound

is always +1 except in metal hydrides, where it is

1.

For example:

The oxidation number of H in H2O and NH3 is +1.However, the oxidation number of H in sodium

hydride, NaH is 1


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Rule 4:

The oxidation number of oxygen in a

compound is always

2 except in peroxides.

For example:

The oxidation number of O in H2O and MgOis 2 . However, the oxidation number of O inhydrogen peroxide, H2O2 is 1


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Rule 5:

The oxidation number of fluorine in all its

compound is 1.

The oxidation number of other halogens

(Cl, Br, I) in their compounds is

1 exceptwhen they combine with more electronegative

elements such as oxygen or nitrogen.


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Rule 6:

The sum of the oxidation numbers of all the

elements in the formula of a compoundmust

be zero.

The sum of the oxidation numbers of all theelements in the formula of a polyatomic ion

must be equal to the charge of the ion.


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Calculation of Oxidation Number, ON

Example 1:Determine the oxidation number of nitrogen in NH3.

Assume that the oxidation number of nitrogen is X

The ON of H in NH3 is +1 (rule 3)The sum of ON of all atoms = 0 (rule 6)

Thus , x + 3(+1) = 0

x + 3 = 0

x =

3

NH3

x +1


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Example 2:Determine the oxidation number of copper in Cu2O

Assume that the oxidation number of copper is X

The ON of O in Cu2O is 2 (rule 4)

The sum of ON of all atoms = 0 (rule 6)Thus , 2x + (2) = 0

2x 2 = 0

x = +1

Cu2O

x

2


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Example 3:Determine the oxidation number of sulphur in SO4

2

Assume that the oxidation number of sulphur is X

The ON of O in SO42 is 2 (rule 4)

The sum of ON of all atoms =

2 (rule 6)Thus , x + 4(2) = 2

x 8 = 2

x = +6

SO42

x

2


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Example 4:Determine the oxidation number of manganese inMnO4

Assume that the oxidation number of manganese is X

The ON of O in MnO4

is

2 (rule 4)The sum of ON of all atoms = 1 (rule 6)

Thus , x + 4(2) = 1

x 8 = 1

x = +7

MnO4

x

2


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Practice A3:

Determine the oxidation number of the underlinedelements in the following compound

a) CO2 b)MgF2 c) H3PO4 d) V2O5

e) CO f) NH4+ g) SO3 h) ClO4

-

i) N2O j) H2O2 k) S2O32 l) CrO4

2

m) Cr2O72 n) Al2O3 o) BrO3

p)VO2

q) PbO22 r) NO3

s) NO2 t) CO3

2

u) HCl v) HClO w) HClO2 x)ClO2

y) HClO3 z)HClO4


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Oxidation Number and IUPAC Nomenclature

Formula of CompoundOxidation Number of

Underlined MetalIUPAC name

FeCl2 +2 Iron(II) chloride

FeCl3 +3 Iron(III) chloride

CuCl +1 Copper(I) chloride

CuSO4 +2 Copper(II) sulphate

Mn(NO3)2 +2 Manganese(II) nitrate

MnO2 +4 Manganese(IV) oxide

K4Fe(CN)6 +2Potassium

hexacyanoferrate (II)

K3Fe(CN)6 +3Potassium

hexacyanoferrate (III)


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Common Names and IUPAC Names for some compound

Molecular

Formula

ON of

metal

Common Name IUPAC name

Na2SO3 +4 Sodium sulphite Sodium sulphate(IV)

Na2SO4 +6 Sodium sulphate Sodium sulphate(VI)

NaNO2 +3 Sodium nitrite Sodium nitrate(III)

NaNO3 +5 Sodium nitrate Sodium nitrate(V)

HNO2 +3 Nitrous acid Nitric(III) acid

HNO3 +5 Nitric acid Nitric (V) acid

H2SO4 +6 Sulphuric acid Sulphuric (VI) acid


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The Changes in Oxidation Number

An increase in oxidation numberindicates Oxidation

A decrease in oxidation number

indicates Reduction


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2Mg(s) + O2 (s) 2MgO(s)Oxidation 0 0 +2

2

Number

Decrease in

oxidation number

Increase in

oxidation number

Example 4:

0

+2

-2

Mg2+

MgO2

O2-

Increase in

Oxidation

Number

(Oxidation) decrease in

Oxidation

Number

(Reduction)


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Explanation

The oxidation number of Magnesium

increases from 0 to +2.

Magnesium undergoes oxidation to

magnesium ion

The oxidation number of Oxygen decrease

from 0 to 2

Oxygen undergoes reduction to oxide ion

Magnesium acts as reducing agent

Oxygen acts as oxidizing agent


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Practice A4

a) 2H2 + O2

2H2Ob) 2Na + Br2 2NaBr

c) Pb + 2Ag+Pb2+ + 2Ag

d) Zn + 2HCl ZnCl2 + H2

Explain the above redox reactions based on thechanges in oxidation number. Your explanationshould includes:

i) oxidized and reduced substance in eachreaction. Give reason for your answer.

ii) oxidizing agent and reducing agent in eachreaction. State what happens to them


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Summary on the Definition of

Oxidation and Reduction

Oxidation Reduction

Gain of Oxygen Loss of Oxygen

Loss of Hydrogen Gain of Hydrogen

Loss of Electrons Gain of Electron

Increase in Oxidation

Number

Decrease in Oxidation

Number


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Examples of Redox Reaction

1. Combustion

2. Extraction of Metals

3. Corrosion of Metals

4. Electrochemistry ( Reaction happen inElectrolytic Cell and Voltaic Cell)

5. Change of Fe2+ to Fe3+ and vice versa

6. Displacement of Metal from its salt solution

7. Displacement of halogen from its halide solution

8. Transfer of electrons at a distance


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Examples of Non Redox Reaction

Neutralization

Precipitation Reaction


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Change of Fe2+ to Fe3+ & vice versa

Iron metal (Fe) exhibits two oxidation numbers,

i.e. +2 and +3

Fe2+ ion can be easily converted into Fe3+ ion.

Fe3+ ion can also be easily converted into Fe2+ ion.

Fe2+

Fe3+

Loss of

electron

Gain of

electron

oxidation

reduction


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Oxidation of Fe2+ to Fe3+

Procedure:

1. Pour 2cm3 of freshly prepared iron(II) sulphate,FeSO

4solution into a test tube.

2. Using a dropper, add bromine water drop bydrop until no further changes are observed.

3. Warm the test tube gently

4. Add NaOH solution slowly into the test tubeuntil it excess.

5. Record the observation.


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Solution Used Observations

FeSO4

+

Br2

Reddish brown bromine water was

decolourized.Green FeSO4 solution turn brown.

When NaOH solution was added, a brown

precipitate formed.

The precipitate is insoluble in excessNaOH


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Half Equation:

Oxidation : Fe2+(aq) Fe3+(aq) + e-

Reduction : Br2(aq) + 2e- 2Br(aq)

------------------------------------------------------------------------------------

Ionic Equation :

2Fe2+ (aq) + Br2 (aq) 2Fe3+(aq) + 2Br(aq)===========================================================



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Reduction of Fe3+ to Fe2+

Procedure:

1. Pour 2cm3 of iron(III) sulphate, Fe2(SO4)3solution into a test tube.

2. Add half a spatula of zinc powder to thesolution. Shake the mixture until no furtherchanges are observed.

3. Filter the mixture.

4. Add NaOH solution slowly into the filtrate untilin excess.

5. Record the observation.


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Solution Used Observations

Fe2(SO4)3

+Zn

Part of Zn powder dissolved.

Brown Fe2(SO4)3 solution turn green. When

NaOH solution was added to the filtrate, a

green precipitate was formed.

The precipitate is insoluble in water.


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Half Equation:

Oxidation : Zn(s) Zn2+ (aq) + 2e-

Reduction : Fe3+

(aq) + e-

Fe2+

(aq)--------------------------------------------------------------------------

Ionic Equation :

Zn(s) + 2Fe3+

(aq)

Zn2+

(aq) + 2Fe2+

(aq)

============================================


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Displacement of Metals

A metal displacement reaction involves a metal

and the salt solution of another metal.

Displacement Reaction took place if any of theseobservation is obtained:

a deposition of solid occurs at the bottom of the test

tube.

a change in colour of the salt solution a decrease in the amount or size of the metal used

Test tube becomes hotter


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A more electropositive

metal can displace a less

electropositive metal from

its aqueous salt solution.

A less electropositive

metal cannot displace a

more electropositive metal

from its aqueous salt

solution


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A more electropositive

metal is located at higher

position in the

electrochemical series, ES.

A less electropositive metal

is located at lower position

in the electrochemicalseries, ES.

K

NaCa

Mg

Al

Zn

Fe

Sn

Pb

Cu

Hg

Ag

More

electropositive

Less

electropositive


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Displacement of MetalsExamples:

Fe(s) + CuSO4(aq) Cu(s) + FeSO4(aq)

Cu(s) + 2AgNO3(aq) 2Ag(s) + Cu(NO3)2(aq)

Mg(s) + FeSO4(aq) Fe(s) + MgSO4(aq)

Cu(s) + FeSO4(aq) No reaction

Zn(s) + MgSO4(aq) No reaction


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Zinc displaces copper metal from copper(II) sulphate solution

Zn(s) + CuSO4(aq) Cu(s) + ZnSO4(aq)

Half Equation: Zn(s) Zn2+(aq) + 2e- (Oxidation)

Cu2+(aq) + 2e- Cu(s) (Reduction)

Overall ionic equation:

Zn(s) + Cu2+(aq) Cu(s) + Zn2+ (aq)


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Observation:

(a) Brown copper metal deposited

(b) The colour of the solution changes from

blue to colourless.

(c) The temperature of the mixture increases.

( all displacement reaction are exothermic)


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Displacement of Halogen

A more reactive halogen can displace a less

reactive halogen from its aqueous halide saltsolution.

A less reactive halogen cannot displace a more

reactive halogen from its aqueous halide saltsolution.

Cl, Br, I

Less reactiveMore reactive


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Colour of halogen

HalogenColour in

aqueous solution

Colour in

1,1,1-trichlorethane

Chlorine Pale yellow Colourless

Bromine Brown Brown

Iodine Brown Purple


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Displacement of halogen

Procedure:

1. 1cm3 of aqueous potassium iodide solution, 1cm3 ofbromine water and 1cm3 of 1,1,1-trichloroethane are addedinto a test tube, labelled A. The mixture is shaken.

2. Step 1 is repeated by adding 1cm3 of aqueous potassiumbromide solution, 1cm3 of chlorine water and 1cm3 of1,1,1-trichloroethane are added into a test tube, labelled B.The mixture is shaken.

Di l f h l


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Displacement of halogen

Result:

Test tubeColour of

CH3CCl3Inference

A Purple Iodinedisplaced

B Brown Brominedisplaced

Di i


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Discussion

Test Tube A

Cl2(aq) + 2KBr(aq) Br2(aq) + 2KCl(aq)

Cl2(aq) + 2Br (aq) Br2(aq) + 2Cl

(aq)

ON : 0

1 0

1

Bromine, Br2 dissolves in CH3CCl3 to give a brown colour

Chlorine, Cl2 is reduced. Reducing agent are the Brions.

Bromide ions are oxidized. Oxidizing agent is Cl2.

Di i


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Discussion

Test Tube B

Br2(aq) + 2KI(aq) I2(aq) + 2KBr(aq)

Br2(aq) + 2I (aq) I2(aq) + 2Br

(aq)

ON : 0

1 0

1

Iodine, I2 dissolves in CH3CCl3 to give a purple colour

Bromine, Br2 is reduced. Reducing agent are the Iions.

Iodide ions are oxidized. Oxidizing agent is Br2.

T f f El Di I


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Transfer of Electron at a Distance I

At electrode X

Iodide ions lose electron and are oxidized to brown iodine.

2I(aq) I2(aq) + 2e-

ON: -1 0 (oxidation)

Dilute sulphuric acid

Th l f h l i h f l l b


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The colour of the solution changes from colourless to brown.

The e- released by the iodide ion flow from electrode X to

electrode Y along the connecting wires.

At electrode Y

The bromine molecules surrounding the electrode Y accept

the e- and are reduced to bromide ions.

Br2(aq) + 2e- 2Br(aq)

ON: 0 -1 (Reduction)

The colour of the solution changes from brown to colourless.


Br2(aq) + 2I(aq) 2Br(aq) + I2(aq)

oxidant reductant

T f f El t t Di t II


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Transfer of Electron at a Distance II

Dilute sulphuric acid

At electrode X

Each Iron(II) ion loses an electron and is oxidized to

brown iron(III) ion. Fe2+ (aq) Fe3+ (aq) + e-

ON: +2 +3 (oxidation)

Th l f h l i h f b


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The colour of the solution changes from green to brown.

The e- released by the iron(II) ion flow from electrode X to


At electrode Y

The bromine molecules surrounding the electrode Y accept

the e- and are reduced to bromide ions.

Br2(aq) + 2e- 2Br(aq)

ON: 0 -1 (Reduction)

The colour of the solution changes from brown to colourless.


Br2(aq) + Fe2+ (aq)2Br(aq) + Fe3+ (aq)

oxidant reductant

T f f El t t Di t III


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Transfer of Electron at a Distance III

At electrode X

Each Iron(II) ion loses an electron and is oxidized to

brown iron(III) ion. Fe2+ (aq) Fe3+ (aq) + e-

ON: +2 +3 (oxidation)

X Y

Th l f th l ti h f t ll /


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The colour of the solution changes from green to yellow/

brown.

The e- released by the iron(II) ion flow from electrode X to


At electrode Y

The manganate(VII) ion, MnO4 gathered at the electrode Y

accept the e- and are reduced to manganese(II) ion, Mn2+.

MnO4 (aq) + 8H+ (aq) + 5e- Mn2+ (aq) + 4H2O(l)

ON: +7 +2 (Reduction)

The colour of the solution changes from purple to colourless.


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anode 5[Fe2+ (aq) Fe3+ (aq) + e-]

cathode MnO4 (aq) + 8H+ (aq) + 5e- Mn2+ (aq) + 4H2O(l)

5Fe2+

(aq) + MnO4

(aq) + 8H+

(aq)

5Fe3+

(aq)+ Mn2+

(aq) + 4H2O(l)

Substance oxidized : iron(II) ion, Fe2+

Substance reduced : manganese(VII) ion, MnO4

Oxidizing agent : manganese(VII) ion, MnO4

Reducing agent : iron(II) ion, Fe2+

P i A5


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Practice A5

The figure above shows an experiment on the transfer of electron at adistance

a) Identify the

i) oxidizing agent ii) reducing agent

iii) positive electrode iv) negative electrode

b) Explain the changes at the

i) negative electrode ii) positive electrode

A B

K2Cr2O7 (aq)

+

H2SO4(aq)

FeSO4(aq)

H2SO4(aq)

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3A Redox Reaction