chapter 10 electrochemistry text book exercise

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CHAPTER 10

ELECTROCHEMISTRY TEXT BOOK EXERCISE

Q1. Multiple choice questions.

(i) The cathode reaction in the electrolysis of dill. H2SO4 with Pt

electrode is

(a) Reduction (b) Oxidation

(c) Both oxidation and reduction(d) Neither oxidation nor reduction

(ii) Which of the following statement is correct about galvanic cell?

(a) Anode is negatively charged

(b) Reduction occurs at anode

(c) Cathode is positively charged

(d) Reduction occurs at cathode

(iii) Stronger the oxidizing agent, greater is the(a) Oxidation potential

(b) Reduction potential

(c) Redox potential

(d) EMF of cell

(iv) If the slat bridge is not used between two half cells, then the

voltage

(a) Decrease rapidly (b) Decrease slowly

(c) Does not change (d) Drops to zero

(v) If a strip of Cu metal is placed in a solution of FeSO4.

(a) Cu will be precipitated out

(b) Fe is precipitated out

(c) Cu and Fe both dissolve



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(d) No reaction takes place

Ans. i) a ii) d iii) b iv) d v) d

Q2. Fill in the blanks with suitable words.(i) The oxidation number of O atom is _____in OF2 and is _____in

H2O2.

(ii) Conductivity of metallic conductors is due to the flow of

_____while that of electrolytes is due to flow of ______.

(iii) Reaction taking place at the ______is termed as oxidation, and at

the _____ is called reduction.

(iv) _____is setup when a metal is dipped in its own ions.

(v) Cu metal _____the Cu-cathode when electrolysis is performed for

CuSO4 solution with Cu-cathodes.

(vi) The reduction potential of Zn is _____volts and its oxidation

potential is ______volts.

(vii) In a fuel cell ______react together in the presence of ______.

Ans. i) +2, -1 ii) electrons. Ions

iii) anode, cathode iv) Equilibrium

v) deposits on vi) -0.76, +0.76

vii) H2 and O2, Pt catalyst

Q3. Indicate TURE or FALSE as the case may be.

(i) In electrolytic conduction electrons flow through the electrolyte.

(ii) In the process of electrolysis, the electrons in the external circuit

flow from cathode to anode.

(iii) Sugar is a non-electrolyte in solid form and when dissolved in

water will allow the passage of an electric current.

(iv) A metal will only allow the passage of an electric current when it

is in cold state.

(v) The electrolyte products of aqueous copper (II) chloride solution



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are copper and chlorine.

(vi) Zinc can displace iron form its solution.

(vii) SHE cats as cathode when connected with Cu electrode.(viii) A voltaic cell produces electrical energy at the expense of

chemical energy.

(ix) Lead storage battery is not reversible cell.

(x) Cr changes its oxidation number when K2 Cr2 O7 is reacted with

HCI.

Ans. i) False ii) False iii) False iv) False

v) True vi) True vii) False viii) Trueix) False x) True

Q4. Describe the electrolysis of molten sodium chloride, and a

concentrated solution of sodium chloride.

See Section 10.2.4

Q5. What is the difference between single electrode potential and

standard electrode potential? How can it be measured? Give its

importance.

Q6. Outline the important applications of electrolysis. Also write the

electrochemical reactions involved therein. Discuss the electrolysis of

Cu SO4 using Cu electrode and AgNO3 solutions using Ag electrode.

See Section 10.2.2

Q7. Describe the construction and working of standard hydrogen

electrode.

See Section 10.3.1

Q8. Is the reaction Fe3+ + Ag EMBED Equation.DSMT4 Fe2+ +

Ag+ spontaneous? If not, write spontaneous reaction involving these

species.

Solution

Fe3+ + Ago EMBED Equation.DSMT4 Fe2+ + Ag+



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In this reaction, Fe is reduced while Ag is oxidized. Therefore, Fe+3

will act as cathode while Ago as anode.

Thus, emf of the cell will beEocell =Eoox + Eored

Eocell = - 0.7994 + (- 0.44)

Eocell = - 0.7994 - 0.44

Eocell = - 1.2394

Since emf of cell is negative, therefore, the cell-reaction is non-

spontaneous.

But if the electrodes are reversed, the cell-reaction becomesspontaneous i.e.

Fe3+ + Ago EMBED Equation.DSMT4 Fe3+ + Ago

Q9. Explain the difference between

(a) Ionization and electrolysis

IONIZATION ELECTROLYSIS 1 The process in which ionic

compounds when fused or dissolved in water split up into charged

particles called ions. 1 The process in which electricity is used to

carry out a non-spontaneous reaction is called electrolysis. 2

Electrodes are not needed 2 Electrodes are required 3 Electricity is

not needed 3 Electricity is required 4 Since there are no electrodes,

therefore, ions do not move towards electrodes 4 Ions moves towards

their respective electrodes 5 After ionization, ions are not discharged

5 Ions are discharged at electrodes to give neutral products.

(b) Electrolytic and Voltaic cell

See Section 10.2.2 and 10.2.5

(c) Conduction through metals and molten electrolytes

CONDUCTION

THROUGH METALS CONDUCTION THROUGH

MOLTEN ELECTROLYTE 1 Electrical conduction takes place due



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to free electrons. 1 Electrical conduction takes place due to ions 2

There in no need to convert metal into molten state. 2 Electrolyte

must be converted into molten state for electrical conduction 3 In thiscase, conductance decreases with increase in temperature. 3 In this

case, conductance increase with increase in temperature. 4 No

chemical reaction occurs during conduction. 4 Chemical reaction

occurs take place during conduction. 5 Chemical composition of

metal is not changed during conduction and no new substance are

produced. 5 Since chemical reactions occur, therefore new substances

are produced. 6 Example:All metals are conductions. e.g. Fe, Pb etc. 6 Example:

Molten salt e.g. NaCI(l) or their aqueous solutions, acids, bases etc.

Q10. Describe a galvanic cell explaining the function of electrodes

and salt bridge.

See Section 10.2.5

Q11. Write comprehensive notes on

Spontaneity of oxidation-reduction reactions.

See Section 10.4.1

(b) Electrolytic conduction

See Section 10.2

(c) Alkaline, silver oxide and Nickle-Cadmium batteries, fuel cells.

See Section 10.5

(d) Lead accumulator, its desirable and undesirable features.

See Section 10.5.1

Q12. Will the reaction be spontaneous for the following set of half

reactions? What will be the value of Ecell?

(i) Cr3+(aq) + 3e- EMBED Equation.DSMT4 Cr(s)

(ii) MnO2(s) + 4H+ + 2e- EMBED Equation.DSMT4 Mn2+(aq) +

2H2O(l)



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Standard reduction potential for reaction

(i) = - 0.7 and for the reaction (ii) = + 1.28 V.

In reaction (i) Cr is reduced from +3 to +2.In reaction (ii) Mn is also reduction reactions; hence these reactions

are not possible in these forms.

However, if reaction (i) is reversed so that Cr is oxidized then the

reaction becomes spontaneous and its emf can be calculated as

(i) Cr(s) EMBED Equation.DSMT4 Cr3+(aq) + 3e- Eoox = + 0.74V

(ii) MnO2(s) + 4H+ + 2e- EMBED Equation.DSMT4 Mn2+(aq) +

2H2O(l) Eo(aq) =+1.28 Vemf of the cell is given by

Eocell =Eooxl + Eored

Eocel = +0.74 + 1.28

Eocell =2.02 V

Q13. Explain the following with reasons

A porous plate or a salt bridge is not required in lead storage cell.

A salt bridge has two main functions

It joins solution of two half cells and thus complete the circuit.

It maintains electrical neutrality of the two half cells as ions can pass

through it.

In lead storage battery, both cathode and anode are dipped in the same

solution. Therefore, excess positive or negative ions are not produced

in the solution. Hence, there is no need of salt bridge.

The standard oxidation potential of Zn is 0.76 V and its reduction

potential is – 0.76 V.

According to the law of conservation of energy, energy can neither be

created nor destroyed. Therefore, if standard oxidation potential of Zn

is 0.76 V, then its potential for reverse process, i.e. standard reduction

potential will also be same but with positive sign. Thus



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Zn EMBED Equation.DSMT4 Zn2+ + 2e- Eoox =0.76 V

Zn2+ + 2e- EMBED Equation.DSMT4 Zn Eoox =0.76 V

(C) Na and K can displace hydrogen from acids but Pt, Pd and Cucannot.

Greater the value of reduction potential, Lesser is the ability to loose

electron to from positive ion, Hence weaker is its tendency to displace

H2-.

Metals like Pt, Pb, and Cu have high positive value of reduction

potential. Thus these do not liberate H2.

Metals like Na and K have negative values of reduction potential.Thus, these can liberate H2.

2Na +2HCI EMBED Equation.DSMT4 2NaCI + H2

2K +2HCI EMBED Equation.DSMT4 2KCI + H2

The equilibrium is set up between metal atoms of electrode and ions

of metal in a cell.

When a metal electrode is dipped into the solution of its own ion.

there may be two tendencies

Metal atom from electrode leaves the electron on metal an goes into

solution. this is oxidation process

M EMBED Equation.DSMT4 M+ + e-

Metal ions in solution may take up electrons form the metal electrode

and deposit as atom on electrode. This is reduction process.

M+ e- EMBED Equation.DSMT4 M+

At last, a dynamic equilibrium is established due to same rate of two

processes. Thus no further potential difference is developed.

(e) A salt bridge maintains the electrical neutrality in the cell.

Two half cells are electrically connected by a salt bridge.

Consider a Zn-Cu cell

During reactions of this cell, Zn half cell continuously loose



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electrons.

Thus, in this positive charge is increasing.

Zn EMBED Equation.DSMT4 Zn2+ + 2e-While, Cu half cell continuously receive electrons, thus it goes on

collecting negative charge.

Cu2+ + 2e- EMBED Equation.DSMT4 Cu

Collection of positive charge in Zn electrode half cell and collection

of negative charge in Cu half cell would stop the reaction.

Salt bridge prevents the net accumulation of charges in either beaker.

Thus form negative Cu half cell, negative ions diffuse through the salt bridge into the positive Zn half cell. In this way, salt bridge maintains

the two solution, electrically neutral.

(f) lead accumulator is a chargeable battery.

See Section 10.5.1

(g) Impure Cu can be purified by electrolytic process.

Impure Cu can be made pure in an electrolytic cell. Thick sheets of

impure copper are made anode, while, thin sheets of pure copper are

made cathode in the cell. These sheets are placed in an electrolytic

solution of CuSO4.

When current is passed through the cell, Cu from anode is oxidized to

Cu2+ ions, which go into the solution. from the solution, Cu2+ ions

are reduced to metallic Cu and deposits as pure Cu on cathode. In this

way, impure sheets of Cu (anode) become this, while pure sheets of

pure Cu (cathode) become thick.

The reactions in the cell are

At Anode (oxidation)

Cu EMBED Equation.DSMT4 Cu2+ + 2e-

At Cathode (reduction)

Cu2+ + 2e- EMBED Equation.DSMT4 Cu



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Thus, there is no net reaction in the cell. However, the net result is the

purification of Cu.

(h) S.H.E. acts as anode when connected with the Cu electrode but ascathode with Zn electrode.

See Section 10.3.2

NUMERICAL PROBLEMS (Exercise)

Q14. (c) Calculate the oxidation number of Chromium in the

following.

CrCl3 K2CrO4

Oxidation number of Cl=-1 Oxidation number of K = + 1

Oxidation number of Cr=x Oxidation number of Cr= - 2

Oxidation number of Cr Oxidation number of Cr=x

can be calculated as Thus

For CrCl3 For K2CrO4

x + 3( - 1) =0 2(+1) + x +4(-2)=0

x – 3 =0 x – 6 =0

or x =+3 x =+6

K2Cr2O7 CrO3

Oxidation number of K=+1

Oxidation number of O= -2 Oxidation number of O= -2

Oxidation number of Cr=x



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Thus Oxidation number of Cr=x

Thus

For K2Cr2O7 For CrO3

2(+1)+2x + 7(-2) =0 x +3(-2)= 0

2x – 12 =0 x – 6 =0

Or x=+12/2 =+6 Or x =+6

Cr2O3 Cr2O72-

Oxidation number of O =-2 Oxidation number of O =-2Oxidation number of Cr = x Oxidation number of Cr = x

Thus Thus

For Or2O3 For Cr2O72-

2x+3(-2)=0 2x +7(-2) = -2

2x – 6 = 0 2x – 14= - 2 +14

Or x= +6/2 =+3 Or x = + 12/2 =+6

Cr2(SO4)3

Oxidation number of S = +16

Oxidation number of O = -2

Oxidation number of Cr = x

Thus

For Cr2(SO4)3

2x +3[(+6)+4(-2)]=0



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2x – 6 =0

Or x=6/2 =+3

(d) Calculate the oxidation numbers of the elements underlined in the

following compounds

Na3 PO4 Na2CO3

Oxidation number of Na = +1

Oxidation number of O = -2 Oxidation number of Na = +1 ThusOxidation number of P = x Oxidation number of C = x

Thus

For Na3 PO4 For Na2CO3

3(+1) + x + 4(-2) =0 2(+1) + x + 3 (-2) = 0

x – 5 =0 x – 4 =0

Or x = +5 Or x =+4

Cr2(SO4)3 Ca(ClO3)2

Oxidation number of Cr = +3 Oxidation number of Ca = +

Oxidation number of O = - 2 Oxidation number of O = - 2

Oxidation number of S = x Oxidation number of S = x

Thus Thus

For Cr2(SO4)3 For Ca(ClO3)2

2(+3) +3 [ (x)+4(-2)] =0 (+2) +2 [ (x)+3(-2)] =0



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3x – 18 =0 2 + 2x -12 =0

Or x = 18/3 = +6 Or x=10/2 = +5

K2 MnO4 HNO3

Oxidation number of K = +1 Oxidation number of H = +1

Oxidation number of O = -2 Oxidation number of O = -2

Oxidation number of Mn = x Oxidation number of N = x

Thus Thus

For K2 MnO4 For HNO32(+1) + x + 3(-2) = 0 (+1) + x +3(-2) =0

x – 6 =0 x – 5 =0

Or x = +6 Or x = +15

HPOs

Oxidation number of H = +1

Oxidation number of O = -2

Oxidation number of P = x

Thus

For HPOs

(+1) + x + 3(-2) =0

x - 5 =0

Or x = +5

Try Yourself

H 2O2 , Ca(OCl)2 , NalO2 , Zn(OH)2 , H3PO4,

Q15. (b) Balance the following equations by oxidation number



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method.

PROBLEM

HNO3 + Hl EMBED Equation.DSMT4 NO + l2 + H2O

Identify the elements, which undergo a change in oxidation number

and write their oxidation numbers over the symbols.

+1+5 2(-2) +1 -1 +2 _-2 o


Determine the no. of electrons gained and lost and equate them.

gain of 3e- x 1 =3 e- (reduction )

+5 -1 +2 o


lose of l e- x 3 = 3 e- (oxidation)

Balance loss and gain of electrons by multiplying Hl by 3.

HNO3 + 3Hl EMBED Equation.DSMT4 NO + l2 + H2O

Balance the rest of equation by inspection method.

2HNO3 + 6Hl EMBED Equation.DSMT4 2NO + 3l2 + 4H2O

PROBLEM

Br2 + NaOH EMBED Equation.DSMT4 NaBrO3 + NaBr + H2O

Identify the elements which undergo a change in oxidation number

and write their oxidation number over the symbols.

o +1 -2 + 1 +1 +5 3(-2) +1 -1



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Br2 + NaOH EMBED Equation.DSMT4 NaBrO3 + NaBr + H2O

Since Br2 is involved both in oxidation and reduction, therefore, we

shall write the Br=twice. Then determine the no. of electrons gainedand lost and equate them.

gain of 2(l) e- x 5 = 10e- (reduction )

o +1 +5 3(-2) +1 -1

Br2 + Br2 + NaOH EMBED Equation.DSMT4 NaBrO3 + NaBr +

H2O

loss of 2(5) e- x 1 = 10 e- (reduction)

Balance loss and gain of electrons by multiplying Br2 by 5, in whichoxidation occurs.

5Br2 + Br2 + NaOH EMBED Equation.DSMT4 NaBrO3 + NaBr

+H2O


5Br2 + Br2 +12 NaOH EMBED Equation.DSMT4 2NaBrO3 +

10NaBr + 6H2O

Or 6Br2 +12 NaOH EMBED Equation.DSMT4 2NaBrO3 + 10NaBr

+ 6H2O

Or 3Br2 +6 NaOH EMBED Equation.DSMT4 NaBrO3 + 5NaBr +

3H2O

PROBLEM

Zn + HNO3 EMBED Equation.DSMT4 Zn(NO3) + NO +H2O



0 +1+5 3(-2) +2+5 +2 -2

Zn + HNO3 EMBED Equation.DSMT4 Zn(NO3) + NO +H2O

In this eq. N is reduced from +5 in HNO3 to +2 in NO. But it’s



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oxidation state is not change in Cu(NO3)2. therefore, write HNO3

twice and determine the number of electrons gained and lost and

equate them.

gain of 3 e- x 2 = 6 e- (reduction )

0 +5 +2 +5 +2 -2

HNO3 +Zn + HNO3 EMBED Equation.DSMT4 Zn(NO3) + NO

+H2O

loss of 2 e- x 3 = 6 e- (oxidation )Use the multiplier obtained above to balance loss and gain of

electrons. Thus Cu is multiplied by 3 and HNO3 by 2.

HNO3 +3Zn + 2HNO3 EMBED Equation.DSMT4 Zn (NO3) + NO

+H2O


6HNO3 +3Zn + 2HNO3 EMBED Equation.DSMT4 3Zn(NO3) +

2NO +4H2O

Or

3Zn + 8HNO3 EMBED Equation.DSMT4 3Zn(NO3) + 2NO +4H2O

PROBLEM

+4 2(-2) + -1 +2 2(-1) 0

MnO2 HCI EMBED Equation.DSMT4 MnCl2 + H2O + Cl2

Identify eq. Cl is reduced from – 1 in HCl to 0 in Cl2. But its

oxidation state is not changed in MnCl2. therefore, write HCl twice

and determine the no. of electrons gained and lost and equate them.




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+4 -1 +2 2(-1) 0

HCI+MnO2 HCI EMBED Equation.DSMT4 MnCl2 + H2O + Cl2

loss of 1 e- x 2 = 2 e- (oxidation )Use the multiplier obtained above to balance loss and gain of

electrons.

HCI+MnO2 2HCI EMBED Equation.DSMT4 MnCl2 + H2O + Cl2


2HCI+MnO2 2HCI EMBED Equation.DSMT4 MnCl2 + 2H2O +

Cl2

OrMnO2 + 4HCI EMBED Equation.DSMT4 MnCl2 + 2H2O + Cl2

PROBLEM

FeSO4 + K2Cr2O7 + H2SO4 EMBED Equation.DSMT4 Fe(SO4) +

Cr2(SO4)3 + K2SO4 + H2O


and write their oxidation numbers over the symbols.

+2 +6 +3 +3


Cr2(SO4)3 + K2SO4 + H2O

Determine the no of electrons gained and lost and equate them.

gain of 2(3) e- x 1 = 6 e- (reduction )

+2 +6 +3 +3


Cr2(SO4)3 + K2SO4 + H2O



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loss of 1(2) e- x 3 = 6 e- (oxidation )

Use the multiplier obtained above to balance loss and gain of electrons

6FeSO4 + K2Cr2O7 + H2SO4 EMBED Equation.DSMT4 3Fe (SO4)+ Cr2 (SO4)3 + K2SO4 + H2O


6FeSO4 + K2Cr2O7 + 7H2SO4 EMBED Equation.DSMT4 3Fe(SO4)

+ Cr2(SO4)3 + K2SO4 + 7H2O

PROBLEMCu + H2SO4 CuSO4 + SO2 + H2O



0 +6 +2 +4 2((-2)

Cu + H2SO4 EMBED Equation.DSMT4 CuSO4 + SO2 + H2O

In this equation S is reduced from +6 in H2SO4 to +4 in SO2. But it’s

oxidation state is not changed in CuSO4. Therefore, write H2SO4

twice and determine the no. of electrons gained and lost equate them.


0 +6 +2 +4

H2SO4+ Cu + H2SO4 EMBED Equation.DSMT4 CuSO4 + SO2 +

H2O

loss of 2 e- x 1 = 2 e- (oxidation )

Use the multiplier obtained above to balance loss and gain of

electrons.


H2O





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2H2O

Or

Cu + 2H2SO4 EMBED Equation.DSMT4 CuSO4 + SO2 + 2H2O

PROBLEM

H2SO4+ Hl EMBED Equation.DSMT4 SO4 + l2 + H2O


and wirte their oxidation number over the symbols.

+6 +1 -1 +4 2(-2) 0H2SO4+ Hl EMBED Equation.DSMT4 SO4 + l2 + H2O

Determine the no. of electrons gained and lost and equate them.


H2SO4+ Hl EMBED Equation.DSMT4 SO4 + l2 + H2O



H2SO4+ 2Hl EMBED Equation.DSMT4 SO4 + l2 + H2O


H2SO4+ 2Hl EMBED Equation.DSMT4 SO4 + l2 + 2H2O

PROBLEM

NaCl + MnO2 + H2SO4 EMBED Equation.DSMT4 Na2SO4 +

MnSO4 + H2O + Cl2



+1 -1 +4 2(-2) +2 0



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MnSO4 + H2O + Cl2

Determine the number of electrons gained and lost and equate them.gain of 2 e- x 1 = 2 e- (reduction )

-1 +4 +2 0


MnSO4 + H2O + Cl2



2NaCl + MnO2 + H2SO4 EMBED Equation.DSMT4 Na2SO4 +MnSO4 + H2O + Cl2


2NaCl + MnO2 + H2SO4 EMBED Equation.DSMT4 Na2SO4 +

MnSO4 + H2O + Cl2

PROBLEM

K2Cr2O7 + HCl EMBED Equation.DSMT4 KCl + CrCl3 + H2O +

Cl2



+6 - 1 +3 0

K2Cr2O7 + HCl EMBED Equation.DSMT4 KCl + CrCl3 + H2O +

Cl2

In this eq. Cl is oxidation from -1 in HCl to 0 in Cl2. But it’s oxidation

state is not change in KCl / or CrCl3. therefore, write HCl twice and

determine the no. of electrons gained and lost and equate them.

gain of 2(3) =6 e- x 1 = 3 e- (reduction )

+6 - 1 +3 0



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HCl+ K2Cr2O7 + HCl EMBED Equation.DSMT4 KCl + CrCl3 +

H2O + Cl2

loss of 1 e- x 6 = 6 e- (oxidation )Use the multiplier obtained above to balance loss and gain of electrons

HCl+ K2Cr2O7 + 6HCl EMBED Equation.DSMT4 KCl + CrCl3 +

H2O + Cl2


8HCl+ K2Cr2O7 + 6HCl EMBED Equation.DSMT4 2KCl +2CrCl3+ 7H2O + 3Cl2

Or

K2Cr2O7 + 14HCl EMBED Equation.DSMT4 2KCl + 2CrCl3 +

7H2O + 3Cl2

Q16. (b) Balance the following equation by ion-election method

PROBLEM

Sn2+ + Fe3+ EMBED Equation.DSMT4 Sn4+ + Fe2+

Identify the elements, which undergo oxidation and reduction and split

up the reaction into oxidation and reduction half reaction.

Fe3+ EMBED Equation.DSMT4 Fe2+

Sn2+ EMBED Equation.DSMT4 Sn4+

Write down the number of electrons gained and lost in each half

reaction

Fe3+ +le- EMBED Equation.DSMT4 Fe2+ (reduction half reaction)

______(1)



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Sn2+ EMBED Equation.DSMT4 Sn4+ + 2e- (oxidation half

reaction) ______(2)

Equate the total number of electrons gained and lost by multiplyingeq. (1) by 2, and then add them.

2Fe3+ +2e- EMBED Equation.DSMT4 2Fe2+

Sn2+ EMBED Equation.DSMT4 Sn4+ + 2e-

Sn2+ +2Fe3+ EMBED Equation.DSMT4 Sn4+ +2Fe2+

PROBLEMH+ + Cl- + Cr2O72- EMBED Equation.DSMT4 Cr3++ Cl2


up the reaction into oxidation and reduction half reactions.

Cr2O72- EMBED Equation.DSMT4 Cr3+ (reduction half reaction)

2Cl- EMBED Equation.DSMT4 Cl2 (oxidation half reaction)

Balance oxygen by adding H2O.

Cr2O72- EMBED Equation.DSMT4 Cr3+ + 7H2O

2Cl- EMBED Equation.DSMT4 Cl2

Balance hydrogen by adding H+ ions.

14H+ + Cr2O72- EMBED Equation.DSMT4 2Cr3+ + 7H2O

2Cl- EMBED Equation.DSMT4 Cl2


reaction.

14H+ + Cr2O72- + 6e- EMBED Equation.DSMT4 2Cr3+ + 7H2O

_________(1)

2Cl- EMBED Equation.DSMT4 Cl2 +2e- _________(2)

Equate the total number of electrons gained and lost by multiplying

eq.(2) by 3. And then add the two half reactions.

14H+ + Cr2O72- + 6e- EMBED Equation.DSMT4 2Cr3+ + 7H2O



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6Cl- EMBED Equation.DSMT4 3Cl2 + 6e-

Cr2O72- +14H+ + 6e- EMBED Equation.DSMT4 2Cr3+ +NO2+

7H2O

PROBLEM (Acidic Media)

Cu + NO3-1 + H+ EMBED Equation.DSMT4 Cu+2 + NO2+ + H2O



NO3-1 EMBED Equation.DSMT4 NO2 (reduction half reaction)Cu EMBED Equation.DSMT4 Cu2+ (oxidation half reaction)


NO3-1 EMBED Equation.DSMT4 NO2 +H2O

Cu EMBED Equation.DSMT4 Cu2+


2H+ + NO3-1 EMBED Equation.DSMT4 NO2

Cu EMBED Equation.DSMT4 Cu2+


reaction

4H+ + NO3-1 + le- EMBED Equation.DSMT4 NO2 +H2O

__________(1)

Cu EMBED Equation.DSMT4 Cu2+ + 2e- __________(2)


eq. (1) by 2. And then add the two half reactions.

4H+ + 2NO3-1 + 2e- EMBED Equation.DSMT4 2NO2 +2H2O

Cu EMBED Equation.DSMT4 Cu2+ + 2e-

2NO3-1 +4H+ + Cu EMBED Equation.DSMT4 Cu2+ + 2NO2



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+2H2O


Cr2O72- + H3AsO3 EMBED Equation.DSMT4 Cr3 + H3AsO4



Cr2O72- EMBED Equation.DSMT4 2Cr3+ (reduction half reaction)H3AsO3 EMBED Equation.DSMT4 H3AsO4 (oxidation half

reaction )


Cr2O72- EMBED Equation.DSMT4 2Cr3 + 7H3O

H2O + H3AsO3 EMBED Equation.DSMT4 H3AsO4


14H+ +6e-+ Cr2O72- + EMBED Equation.DSMT4 2Cr3 + 7H3O

________(1)

H2O +3 H3AsO3 EMBED Equation.DSMT4 3 H3AsO3 + 2e- + 2H+

________(2)

Equate th total number of electrons gained and lost by multiplying eq.

(2) by 3. And then add the two half reactions.

14H+ +6e-+ Cr2O72- + EMBED Equation.DSMT4 2Cr3 + 7H3O

3H2O +3 H3AsO3 EMBED Equation.DSMT4 3 H3AsO3 + 6e- + H+

Cr2O72- +8H+ + 3H3AsO3 EMBED Equation.DSMT4 2Cr3 +

3H3AsO4 + 4H2O



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24



MnO4-+ Cr2O72- EMBED Equation.DSMT4 Mn2+ + CO2



MnO4- EMBED Equation.DSMT4 Mn2+ (reduction half reaction)

Cr2O72- EMBED Equation.DSMT4 2CO2 (oxidation half reaction)

Balance oxygen by adding H2O.MnO4- EMBED Equation.DSMT4 Mn2+ + 4H2O

Cr2O72- EMBED Equation.DSMT4 2CO2


8H+ +MnO4- EMBED Equation.DSMT4 Mn2+ + 4H2O

Cr2O72- EMBED Equation.DSMT4 2CO2

Write down th number of electrons gained and lost in each half

reaction

5e-+ 8H+ +MnO4- EMBED Equation.DSMT4 Mn2+ + 4H2O

__________(1)

Cr2O72- EMBED Equation.DSMT4 2CO2 +2e-1 __________(2)

Equate th total number of electrons gained and lost by multiplying eq.

(2) by 5 and eq. (1) by 2. And then add the two half reactions.

10e-+ 16H+ +2MnO4- EMBED Equation.DSMT4 2Mn2+ + 8H2O +

10CO2

5Cr2O72- EMBED Equation.DSMT4 10CO2 +10e-1



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25


5C2O42-+ 16H+ +2MnO4- EMBED Equation.DSMT4 2Mn2+ +

8H2O + 10CO2


Fe2+ + C2O72- EMBED Equation.DSMT4 Cr3+ + Fe3+


up the reactions into oxidation and reduction half reactions.

C2O72- EMBED Equation.DSMT4 Cr3+ (reduction half reaction)

Fe2+ EMBED Equation.DSMT4 Fe3+ (oxidation half reaction)Balance oxygen by adding H2O.

C2O72- EMBED Equation.DSMT4 2Cr3+ + 7H2O.


14H++ C2O72- EMBED Equation.DSMT4 2Cr3+ + 7H2O.

Fe2+ EMBED Equation.DSMT4 Fe3+

Write down th number of electrons gained and lost in each half

reaction

14H+ +6e-+ Cr2O72- EMBED Equation.DSMT4 2Cr3 + 7H3O

________(1)

Fe2+ EMBED Equation.DSMT4 Fe3+ + le- ________(2)



14H+ +6e-+ Cr2O72- EMBED Equation.DSMT4 2Cr3 + 7H3O

6Fe2+ EMBED Equation.DSMT4 6Fe3+ + 6e-

Cr2O72- +14H+ +6e- EMBED Equation.DSMT4 6Fe2++ 2Cr3 +

7H3O



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26



lO3- + AsO33- EMBED Equation.DSMT4 l- + AsO43-


up the reaction into oxidation and reduction and reduction half

reactions.

lO3- EMBED Equation.DSMT4 l- (reduction half reaction)

AsO33- EMBED Equation.DSMT4 AsO43- (oxidation half reaction)

Balance oxygen by adding H2O.lO3- EMBED Equation.DSMT4 l- + 3H2O

H2O+ AsO33- EMBED Equation.DSMT4 AsO43-

Balance oxygen adding H+ ions.

6H+ + lO3- EMBED Equation.DSMT4 l- + 3H2O

H2O+ AsO33- EMBED Equation.DSMT4 AsO43- + 2H+


reaction

6e- + 6H+ + lO3- EMBED Equation.DSMT4 l- + 3H2O

___________(1)

H2O+ AsO33- EMBED Equation.DSMT4 AsO43- + 2H+ + 2e -

___________(2)



6e- + 6H+ + lO3- EMBED Equation.DSMT4 l- + 3H2O

3H2O+ 3AsO33- EMBED Equation.DSMT4 3AsO43- + 6H+ + 6e -



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27


lO3- + 3AsO33 EMBED Equation.DSMT4 3AsO43- +l-


Cr3+ + BiO3 EMBED Equation.DSMT4 Cr2O72- + Bi3+



BiO3 EMBED Equation.DSMT4 Bi3+ (reduction half reaction)

Cr3+ EMBED Equation.DSMT4 Cr2O72- (oxidation half

reaction)Balance oxygen by adding H2O.

BiO3 EMBED Equation.DSMT4 Bi3+ + 3H2O

7H2O + Cr3+ EMBED Equation.DSMT4 Cr2O72-


6H+ + BiO3 EMBED Equation.DSMT4 Bi3+ + 3H2O

7H2O + Cr3+ EMBED Equation.DSMT4 Cr2O72- + 14H+


reaction

2e- + 6H+ + BiO3 EMBED Equation.DSMT4 Bi3+ + 3H2O

________(1)

7H2O + 2Cr3+ EMBED Equation.DSMT4 Cr2O72- + 14H+ __

_______(2)


eq.(1) by 3. And then add the tow half reactions.

6e- + 8H+ + 3BiO3 EMBED Equation.DSMT4 3Bi3+ + 9H2O

7H2O + 2Cr3+ EMBED Equation.DSMT4 Cr2O72- + 14H+

3BiO3 + 2Cr3+ EMBED Equation.DSMT4 3Bi3+ + Cr2O72- +



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28


2H2O


OCl- + S2O32- EMBED Equation.DSMT4 Cl- + S4O62-



OCl- EMBED Equation.DSMT4 Cl- (reduction half reaction)

S2O32- EMBED Equation.DSMT4 S4O62- (oxidation half reaction)

Balance oxygen by adding H2O.OCl- EMBED Equation.DSMT4 Cl- + H2O

2 S2O32- EMBED Equation.DSMT4 S4O62-


OCl- + 2H+ EMBED Equation.DSMT4 Cl- + H2O

2 S2O32- EMBED Equation.DSMT4 S4O62-


reaction

2e- + OCl- + 2H+ EMBED Equation.DSMT4 Cl- + H2O

________(1)

2 S2O32- EMBED Equation.DSMT4 S4O62-+ 2e- ________(2)

Add the two half reactions.

2e- + OCl- + 2H+ EMBED Equation.DSMT4 Cl- + H2O

2 S2O32- EMBED Equation.DSMT4 S4O62-+ 2e-

2 S2O32- + OCl- + 2H+ EMBED Equation.DSMT4 S4O62-+ CO2




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29


MnO4- + C2O42 EMBED Equation.DSMT4 MnO2 + CO2


up the reaction into oxidation and reduction half reactions.MnO4- EMBED Equation.DSMT4 MnO2 (reduction half reaction)

C2O42 EMBED Equation.DSMT4 CO2 (oxidation half reaction)

Add two OH- ions for one each oxygen atom on appropriate side.

MnO4- EMBED Equation.DSMT4 MnO2 + 4 OH-

C2O42 EMBED Equation.DSMT4 CO2

Balance hydrogen by adding H2O ions.

2H2O + MnO4- EMBED Equation.DSMT4 MnO2 + 4 OH-C2O42 EMBED Equation.DSMT4 CO2

Write down the number of electron gained and lost in each half

reaction

3e-+ 2H2O + MnO4- EMBED Equation.DSMT4 MnO2 + 4 OH-

________(1)

C2O42 EMBED Equation.DSMT4 CO2 + 2e- ________(2)


eq. (1) by 2 and eq. (2) by 3. And then add the two half reactions.

6e-+ 4H2O + 2MnO4- EMBED Equation.DSMT4 2MnO2 + 8OH-

3C2O42 EMBED Equation.DSMT4 6CO2 + 6e-

3C2O42-+2MnO4- + 4H2O EMBED Equation.DSMT4 2MnO2 +

8OH-


MnO4- + CN- EMBED Equation.DSMT4 MnO2 + CNO-





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30


MnO4- EMBED Equation.DSMT4 MnO2 (reduction half

reaction)

CN- EMBED Equation.DSMT4 CNO- (oxidation half reaction)Balance hydrogen and oxygen by adding H2O and OH- ions.

2H2O +MnO4- EMBED Equation.DSMT4 MnO2 +4 OH-

_________(1)

CN- +2 OH- EMBED Equation.DSMT4 CNO- + H2O _________(2)


reaction

3e-+ 2H2O + MnO4- EMBED Equation.DSMT4 MnO2 + 4OH- __________(1)

CN- +2OH- EMBED Equation.DSMT4 CNO- + H2O +2e-

_________ __(2)



6e-+ 4H2O + 2MnO4- EMBED Equation.DSMT4 2MnO2 + 8OH-

___________(1)

3CN- + 6OH- EMBED Equation.DSMT4 3CNO- + 3H2O +6e-

_________ __(2)

3CN- + H2O +2MnO4- EMBED Equation.DSMT4 2MnO2+ 6OH-

+3CNO-


H3AsOs + Cr2O72- EMBED Equation.DSMT4 Cr3+ + H3AsO4



Cr2O72- EMBED Equation.DSMT4 Cr3+ (reduction half reaction)

H3AsOs EMBED Equation.DSMT4 H3AsO4 (oxidation half



1st


reaction)


Cr2O72- EMBED Equation.DSMT4 Cr3+ +7 H2OH3AsOs + H2O EMBED Equation.DSMT4 H3AsO4


14H+ +Cr2O72- EMBED Equation.DSMT4 Cr3+ +7 H2O

H3AsOs + H2O EMBED Equation.DSMT4 H3AsO4 + 2H+


reaction.

14H+ + 6e- +Cr2O72- EMBED Equation.DSMT4 Cr3+ +7 H2O _______(1)

H3AsOs + H2O EMBED Equation.DSMT4 H3AsO4 + 2H+ + 2e-

_______(2)



14H+ +6e- +Cr2O72- EMBED Equation.DSMT4 2Cr3+ +7 H2O

________(1)

3H3AsOs + 3H2O EMBED Equation.DSMT4 3H3AsO4 + 6H+ +

6e- ________(2)

Cr2O72- +8H+ + 3H3AsOs EMBED Equation.DSMT4 3H3As

chapter 10 electrochemistry text book exercise

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