34785536-ELECTROCHEMISTRY.pdf

Faculty-Wide Courses 2009-2010 ELECTROCHEMISTRY

Eng. ISLAM IBRAHIM FEKRY Page 1 Mob. 0109790568

Mail. [email protected]

SHEET 3

1- Consider a galvanic cell based on the reaction

Al+++ (aq) + Mg(s) Al(s) + Mg++(aq)

v37.2-

v66.1-

MgMg

AlAl

E

E

Give the balanced cell reaction and calculate E for the cell.

Since a galvanic cell, so that E have to be +ve value

1-Half reactions

A-Oxidation Mg++ (aq) + 2 e- Mg(s) v37.2-

MgMgE

Mg(s) Mg++ (aq) + 2 e- v37.2

MgMgE

B-Reduction Al+++ (aq) + 3 e- Al(s) v66.1-

AlAlE

2-Balance the equation

3Mg(s) 3Mg++ (aq) + 6 e- v37.2

MgMgE

2Al+++ (aq) + 6 e- 2Al(s) v66.1-

AlAlE

3-Net equation 3Mg(s) +2Al+++ (aq) 3Mg++ (aq)+ 2Al(s)

4- Ecell = E(Cathod, Reduction) - E (Anode, Oxidation)

E = (-1.66) (-2.37) = 0.71 v

Ecell = E(Cathod, Reduction) + E (Anode, Oxidation)

E = (-1.66) + (2.37) = 0.71 v




2- Describe completely the following galvanic cell based on the following

half reactions.

Cl2 + 2e- 2 Cl- E=1.36 v

Br2 + 2e- 2 Br- E= 1.09 v

Since a galvanic cell, so that E have to be +ve value

Note:

As he didnt give me the final net reaction so, I will make assumption which of

the two reaction is will be oxidation Anode and which will be reduction Cathode,

on my assumption, I will take care that the net value of the E have to be +ve value

I will assume that the reaction have a E highest value is the reduction

Cathode, and the other is the oxidation anode to ensure that the net value of E is

+ ve value.

1-Half reactions

A-Oxidation Br2 + 2e- 2 Br- E= 1.09 v

2 Br- Br2 + 2e- E= -1.09 v

B-Reduction Cl2+ 2 e- 2 Cl - E= 1.36 v

2-Balance the equation

2 Br- Br2 + 2e- E= -1.09 v

Cl2+ 2 e- 2 Cl - E=1.36 v

3-Net equation 2 Br + Cl2 Br2 + 2 Cl -


E = (1.36) (1.09) = 0.27 v


E = (1.36) + (-1.09) = 0.27 v




3- Predict whether the following reaction would proceed spontaneously as

written at 298 K.

Co(s) + Fe++ (aq) Co++ (aq) + Fe(s)

Given that [Co++] =0.15 M and [Fe++] =0.68 M

v44.0-=E

v28.0-=E

Fe++Fe

C o++C o

1-Half reactions and balance the equation

A-Oxidation Co++ (aq) + 2e- Co(s) E= -0.28 v

Co(s) Co++ (aq) + 2e- E= 0.28 v

B-Reduction Fe++ (aq) + 2 e- Fe(s) E= -0.44 v

2-Net equation Co(s) + Fe++ (aq) Co++ (aq) + Fe(s)


E = (-0.44) (-0.28) = -0.16 v


E = (-0.44) + (0.28) = -0.16 v

Dependence of Cell Potential on Concentration

4- Nernest Equation

)Qlog(

n

0591.0 -E=E

E = -0.16 (0.0591/ 2) * ( log ( [Co++] / [Fe++] )

E = -0.16 (0.02955) * ( log ( [0.15] / [0.68] )

E = -0.16 (0.02955) * ( log ( [0.220588] )

E = -0.16 (0.02955) * ( -0.65641 )

E = -0.16 (-0.019397 )

E = -0.140 v

Since the value of E is ve, so that this is an electrolytic cell




4- Calculate the EMF for the cell based on the following half reactions

VO2 + 2 H+ + e- VO++ + H2O E = 1.00 v

Zn++ + 2 e- Zn E = -0.76 v

Where: T=25C [VO2++] =2.0M

[H+]=0.5M

[VO++] =1.0x10-2M

[Zn++]=1.0x10-1M

See page 839, and 840 in CHEMISTRY ZUMDAHL|ZUMDAHL BOOK

EXAMPLE 18.9




5- Consider the cell described below:

Zn|Zn++ (1.00M) || Cu++ (1.00 M) |Cu

Calculate the cell potential after the reaction has operated long enough for

the [Zn++] to have changed by 0.2 mol/L. (Assume T=25C)

1-Half reactions and make the Balance

A-Oxidation Zn++ + 2e- Zn E= -0.76 v

Zn Zn++ + 2e- E= 0.76 v

B-Reduction Cu+++ 2 e- Cu E= 0.34 v

2-Net equation Zn + Cu++ Cu + Zn++


E = (0.34) (-0.76) = 1.1 v


E = (0.34) + (0.76) = 1.1 v


4- Nernest Equation

)Qlog(

n

0591.0 -E=E

E = 1.1 (0.0591/ 2) * ( log ( [ Zn++ ] / [Cu++ ] )

E = 1.1 (0.02955) * ( log ( [ 200 ] / [1 ] )

E = 1.1 (0.02955) * ( log ( [ 200] )

E = 1.1 (0.02955) * ( log ( [ 2] )

E = 1.1 (8.8 *10-3)

E = 1.091 v




6- Calculate EMF of the following cell if E Zn++Zn = -0.76

Zn(s) |Zn++ (aq, 0.1M) ||Zn++ (aq, 1.0M) |Zn(s)


A-Oxidation Zn++ + 2e- Zn E= -0.76 v

Zn Zn++ + 2e- E= 0.76 v

B-Reduction Zn++ + 2e- Zn E= -0.76 v

2-Net equation ..... .....


E = (-0.76) (-0.76) = ZERO v


E = (-0.76) + (0.76) = ZERO v


4- Nernest Equation

)Qlog(

n

0591.0 -E=E

E = 0 (0.0591/ 2) * ( log ( [ Zn++ 0.1 M] / [Zn++ 1 M] )

E = 0 (0.02955) * ( log ( [ 0.1 ] / [1 ] )

E = 0 (0.02955) * ( log ( [ 0.1] )

E = 0 (0.02955) * ( -1 )

E = 0.02955 v




7- Calculate the [Cu++] / [Zn++] ratio at which the following reaction will

become spontaneous at 25C.

Cu(s) + Zn++ (aq) Cu++ (aq) + Zn(s)

Reaction to become spontaneous it have to be a galvanic cell, so that the E have a + ve

value


A-Oxidation Cu+++ 2 e- Cu E= 0.34 v

Cu Cu+++ 2 e- E= -0.34 v

B-Reduction Zn++ + 2e- Zn E= -0.76 v

2-Net equation Cu + Zn++ Zn + Cu++


E = (-0.76) (0.34) = -1.1 v


E = (-0.76) + (-0.34) = -1.1 v


4- Nernest Equation

)Qlog(

n

0591.0 -E=E

E = -1.1 (0.0591/ 2) * ( log ([Cu++ ] / [ Zn++ ]))

E = -1.1 (0.02955) * ( log ([Cu++ ] / [ Zn++ ]))

E have a + ve value

-1.1 (0.02955) * ( log ([Cu++ ] / [ Zn++ ])) > zero

-1.1 > (0.02955) * ( log ([Cu++ ] / [ Zn++ ]))

-1.1 / 0.02955 > ( log ([Cu++ ] / [ Zn++ ]))

-37.225 > ( log ([Cu++ ] / [ Zn++ ]))

10-37.225 > ([Cu++ ] / [ Zn++ ])




8- Consider the following half reactions

MnO4-(aq) + 8 H+(aq) + 5e- Mn++(aq) + 4 H2O (L) E=1.51

NO3-(aq) + 4 H+(aq) + 3 e- NO(g) + 2 H2O (L) E=0.96

Predict whether NO3- ions will oxidize Mn2+ to MnO4- under standard

state conditions.

1-Half reactions

A-Oxidation Mn++(aq) + 4 H2O (L) MnO4-(aq) + 8 H+(aq) + 5e-

E= -1.51 v

B-Reduction NO3-(aq) + 4 H+(aq) + 3 e- NO(g) + 2 H2O (L)

E=0.96 v

2-Balance the two equation

A-Oxidation 3Mn++(aq) + 12 H2O (L) 3MnO4-(aq) + 24 H+(aq) + 15e-

E= -1.51 v

B-Reduction 5NO3-(aq) + 20 H+(aq) + 15 e- 5NO(g) + 10H2O (L)

E=0.96 v

3-Net equation

3Mn++(aq) + 2 H2O (L)+ 5NO3-(aq) 3MnO4-(aq) + 4 H+(aq)+ 5NO(g)


E = (0.96) (1.51) = -0.55 v


E = (0.96) + (-1.51) = -0.55 v

Since the value of E is ve, so that this is an electrolytic cell




9- How long will it take to plate out 1.0 kg Al from aq Al+++ with a current of

100.0 A? (M.wt. of Al=26.98).

1- Plate out 1.0 kg Al

2- Moles of Al = [Mass/M.wt.Al] = [1.0*1000 gm/ 26.98] = 37.0644 gmole

3-Moles of Electron Al from aq Al+++

1 gmole of Al 3 e-

37.0644 gmole ??? e-

??? e- = 3*37.0644 = 111.193 e-

4- Quantity of charge in coulombs

1 e 96,485 coulombs

111.193 e ??? coulombs

??? coulombs = 96,485*111.193 = 10,728,502.59451 coulombs

5-Current and time

Ampere= Coulombs / sec

100.0 A= 10,728,502.59451 coulombs / ??? sec

??? sec = 10,728,502.59451 coulombs / 100 A

??? sec = 10,728,5.02 sec

Time 30 hr




10- What mass of Co from aq Co++ can be produced in 1.0 h with a current of

15 A? (M.wt. of Co= 59).

See page 866, in CHEMISTRY ZUMDAHL|ZUMDAHL BOOK QUESTION

NO.91.a

11-An unknown metal M is electrolyzed. It took 74.1 s for a current of 2.00 A

to plate out 0.107 g of the metal from a solution containing M(NO3)3.

Identify the metal.

See page 866, in CHEMISTRY ZUMDAHL|ZUMDAHL BOOK QUESTION

NO.93

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34785536-ELECTROCHEMISTRY.pdf