What are the 2 kinds of cells?

p. 731

This image is from http://www.westga.edu/~chem/courses/chem410/3510_06/sld006.htm

What is oxidation, reduction anode cathode?
                                                         (See OH # 2 & 2-1)

What is a volt?
J/Coulomb

What is a coulomb (C)?
Charge                                                 1 Mole e- = 96, 500 C= F
                                                                                                   (A Faraday) ¿

What is an Ampere?
                                    C/s
                                 Q= I*t
                   Charge = Current * Time

I. Voltaic Cell: A spontaneous reaction used to produce electric energy

Zn/Zn²⁺ // Cu²⁺/Cu

Also from http://www.westga.edu/~chem/courses/chem410/3510_06/sld006.htm

See pg.734 Know by heart!
See OH #3

Salt bridge allows current to flow but prevents contact between Zn and Cu²⁺, which would short-circuit the cell.

See Fig. 21.3 p. 735 OH 5.

Read 5 items p. 737
What is S.H.E.? Why have it?
            Can’t measure single ½ pot.

Standard Hydrogen Electron at 1Atm, 1M[H⁺], 25° C

2H⁺ + 2e^- -> H_2(g) E⁰_red = ZERO, "0", Volts

This can also be found at http://www.westga.edu/~chem/courses/chem410/3510_06/sld007.htm

See table 21.1 (OH 6)

Those at the top really don’t want to happen as written. Those near the bottom (most Å ) do. If you want a E⁰_ox, switch sign

Zn ® Zn²⁺ + 2e^-                     E⁰_ox=0.762

Voltage is electrical pressure and just as with water the pressure does not depend upon how much water is present, voltage is independent of how much. So if you multiply and equation by 2, "E" it is the same! Voltage (EMF or potential difference) depends on P.E. which is determined by bonding arrangements of the atoms. This is the "Driving Force." What does depend upon how much is current?

B. Relative strengths of oxidizing and reducing agents

1. Oxidizing agents (left column, Table 21.1). The larger (more positive) the value of E⁰_red the stronger the oxidizing agent.

 Oxidizing agents become stronger moving down left column

    2.    Reducing agents (right column, Table 21.1). The larger the value of E°, the       stronger the reducing agent.

Reducing agents become weaker moving down right column

Any substance on the left will react with any substance on the right that is above it. Just because you write a reaction and say it goes says nothing about the rate.

C. Calculation of standard cell voltage

E⁰_tot = E⁰_ox + E⁰_red

Cl_2(g) + 2Br^-_(aq) ® 2Cl^-_(aq) + Br_2(l)

E⁰_tot = E⁰_red Cl_2(aq) + E⁰_ox Br^-_(aq)

DG°(J) = -RTlnK

So ln K = nFE°_tot
                     RT

If E°_tot +, DG° is -, K > 1

For Zn + Sn²⁺ à Sn + Zn²⁺

Fine ‘em! E°, DG°, K

E°_tot = -(-0.762) + (-0.141) = 0.621 V

DG° = -2 x (96.5)(0.621) = -120. KJ

ln K = - DG = +120. KJ x 1000 J/kJ
                RT     8.31J/moleK(298K)

K = 1.1 x 10²¹

I found out what to do if K > 10⁹⁹

Ln K = 250 

e^lnK = e²⁵⁰ = K

Effect of concentration on voltage (EMF) (Non standard conditions)

Reaction 1 & 2 p. 749

1. V up con. reactants up or products down
2. V down if con reactants down or products up

For Daniel cell Zn + Cu²⁺ à Zn²⁺ + Cu

Q = [Zn²⁺]          E down as Q up
       --------
        [Cu²⁺]

From "Le Chat" as products [ ] go up we know the system would oppose this so E_net goes down (E_net < E°_net)

p. 750 Nernst Equ.

aA + bB « cC + dD

The Nernst Equ. Is related to Le Chatelier’s principle in that they both relate to what happens to reaction tendencies when concentrations are changed. When E > 0 reactions are favored to go as written. However Nernst is QUANtitative, LeChat QUALitative.

E_net = E°_net - RT lnQ
                          nF

Q = [C]^c[D]^d
       -----------
        [A]^a[B]^b

Q is like equ. Expression BUT at equ. Opposing reaction rates are =; e^- would flow through the wire in opposite directions at an = rate and E_net = 0 therefore Q is symbol for the word reaction "Quotient" and has the same form as the equilibrium constant, K but it applies to any reaction not just equilibrium.

DG = DG° + RT ln Q

If R = 8.31 J/mole*K ; T = 298K

F = Value of Faraday 96,485 J/volt

n = # of e^- transferred

RT = 0.0257
 F

E_net = E°_net – 0.0257 lnQ
                               n

See p. 750 bottom on Q

There are 2 conditions you can get E easily

1. Standard conditions – what is Q? So what is E?

Q = 1 => lnQ = 0 therefore E = E°

2. At equilibrium what is Q, E?

At Equ Q = K but E_cell = 0

A "Dead" battery has reached Equ – "Old chemist never die, they just reach equilibrium."

E°= RT ln K               So we have a neat way of getting K by measuring E°
       nF 

Example: Fe²⁺ + H⁺ + MnO₄^- à Fe³⁺ + Mn²⁺

Given: T = 25°C, [Fe²⁺] = 0.50M

[MnO₄^-] = 0.100M, [H⁺] = 1.00M, [Mn²⁺] = 2x10^-2M

[Fe³⁺] = 3 x 10^-1M

Find: E°_net, E_net & DG

                              5Fe²⁺ à 5Fe³⁺ + 5e^-                     -0.77 V
         8H⁺ + MnO₄^- + 5e^- à Mn²⁺ + 4H₂O                   1.51 V
     5Fe²⁺ + 8H⁺ + MnO₄^- à 5Fe³⁺ + Mn²⁺ + 4H₂O    +0.74 V

= 0.74 – (-0.0021) = 0.76 V

DG = -NFE = -5(96,485 J/V mole)(0.76V)

= -370 kJ/mole

Zn_(s) + 2H⁺_(aq) à Zn²⁺_(aq) + H_2(g)

Suppose Zn²⁺ = 1 M

H₂ = 1ATM [Note ATM’s]

Fine [H⁺] & pH

If E measures 0.200 V [Tell me why!]

USE Nernst

E = E° - 0.0257 ln [H₂][Zn²⁺]
                      n            [H⁺]²

So we need E°

0.762V

0.200V=0.762V-(0.0257)ln(1/[H⁺]²)

-0.562V=-0.0129ln(1/[H⁺]²)

-0.562V=0.0129ln[H⁺]²

ln[H⁺]²=-(0.562/0.0129)=-43.6

2ln[H⁺]=-43.6Þ ln[H⁺]=-21.8

[H⁺]=3.41x 10^-10

^{pH = 9.468 (Note: 4 sig. figs)}

Electrolytic Cells

Electrolysis: By supplying electrical energy we can force a non-spontaneous reaction to go.

If we try to pass an electric current through NaCl(s)-nothing, a good insulator, but if we melt it, ions are free to move.

A.  Molten(fused) salts MX

      

M⁺ + e^-® M




X^- ® X + e^-



See OH(18)

http://www.scifun.chem.wisc.edu/chemweek/Aluminum/Aluminum.html

B.  Electrolysis of Aqueous Solutions

To do this right, we should list ALL reactions that can occur at the electrodes. This needs to include water itself being oxidized or reduced.

Ox: 2H₂O® 4e^- + 4H⁺ +O₂ E° =-1.23V

Red: 2H₂O +2e^-® H₂ + 2OH^- E° =-.83V

@Electrolysis of NaCl_(aq) (IM) Do it

Write down what can happen on each electrode (there is one more factor we should look at on whether we are in acid, base, neutral…but we will ignore consideration here.

At Anode:

2Cl^-_(aq) ® Cl_2(g) +2e^-                      E° = -1.36V

2H₂O ® 4e^- + 4H⁺ + O₂                 E° = -1.23V

At Cathode:

Na_(aq) + e ® Na_(g)                               E° = -2.71V

2H₂O +2e^- ® H_2(g) + 2OH^-_(aq)        E° = -.83V

We would pick water in both cases because H₂O is more easily reduced and oxidized. Choose the one with the most favorable (most +, least -) E° value. BUT! WE smell Cl₂ ! What’s going on?!

Note: The voltages are close to each other so both may come off. Really the evolution of O₂ on graphite has a rather high overvoltage, so that Cl_2(g) comes off. O₂ is predicted by E° , D G but the rate is slow, mechanism considered the different between E (theoretical) and E(real) [not real, well understood]

Factors responsible for difference:

2. Surface conditions at interface

The overvoltage for 2H⁺ + 2e^- ® H₂ is only 0.02V while 2H₂O ® O₂ + 4H⁺ + 4e^- has a large overvoltage. So Cl₂ wins!

So the overall reaction becomes

2Cl^–_(aq) ® Cl_2(g) + 2e^–                                                       –1.36 V
2H₂O_(l) + 2e^– ® 2H_2(g) + 2OH^–_(aq)                                     –0.83 V

2Cl^–_(aq) +2H₂O_(l) ® 2H_2(g) + 2OH^–_(aq) + Cl_2(g)        E_NET = –2.19 V

So to make happen we need a minimum 2.19 V (not counting overvoltage)

Discuss, sketch what happens in the electrolysis of 1M HBr. What is E⁰_NET?



Anode:                                                     E⁰

2Br^–_(aq) ® Br_2(g) + 2e^–                           -1.08V
2H₂O_(l) ® 4e^– + 4H⁺_(aq) + O_2(g)             -1.23V

Cathode:

2H⁺_(aq) + 2e^– ® H_2(g)                             0.00V
2H₂O_(l) + 2e^– ® H_2(g) + 2OH^–_(aq)             -0.83V

---

2Br^–_(aq) + 2H⁺_(aq) ® Br_2(g) + H_2(g)         E⁰_NET = -1.08V

E_REAL > E⁰_NET in an absolute sense

Discuss, sketch what happens in the electrolysis of CuCl₂ Solution (1M) using Graphite Electrodes

Anode:                                                     E⁰

2Cl^–_(aq) ® Cl_2(g) + 2e^–                            -1.36V

2H₂O_(l) ® 4e^– + 4H⁺_(aq) + O_2(g)             -1.23V

Cathode:

Cu⁺²_(aq) + 2e^– ® Cu_(s)                            +0.34V

2H₂O_(l) + 2e^– ® H_2(g) + 2OH^–_(aq)           -0.83V

---

Cu⁺²_(aq) + 2Cl^–_{(aq) ®} Cl_2(g) + Cu_(s) E⁰ = -1.02V

Electrolysis involving active electrodes.

Here the electrodes themselves can undergo a reaction. Used in plating cells.

Discuss, sketch what happens in the electrolysis of CuCl₂ Solution (1M) using copper electrodes.

Anode: 2Cl–(aq) ® Cl2(g) + 2e– 2H2O(l) ® 4e– + 4H+(aq) + O2(g) Cu(s) ® Cu+2(aq) + 2e– Cathode: Cu+2 + 2e- ® Cu 2H2O + 2e- ® H2 + 2OH-

E0 -1.36V -1.23V +0.34V   +0.34V -0.83V

One electrode loses mass (anode), the other gains same mass (cathode). Atoms just transferred. If you have a mixture of metals, the one with the largest E° will plate out first.

Faraday found the amount of chemical change that occurs in an electrolysis reaction is related to the amount of charge that goes through the system.

FARADAY= The amount of charge= to 1 mole of e^-

So, 1F = 1 mole e^-

1 coulomb= 1 ampere x 1 second

Experimentally* 1F = 96,500C = 1 mole e^-

Charge ® Q = I x C (current x time)

C = A x S

Problem: In a copper plating cell, how many grams of Cu will be deposited from a CuSO₄ solution by a current of 1.5A for 4.0 hours?

1. First write what’s given and the equation:

2e^- + Cu⁺²_(aq) ® Cu_(s)

T = 4.0h x 60min./1hr x 60sec/1min. = 14,000 sec.

I = 1.5 A

Find: Mass of Cu

2. q = I x C = 1.5 A x 14,000 sec. = 22,000C
3. 22,000C x 1mol e^-/96,500C x 63.5g Cu/2mol e^- = 7.0g Cu

Problem: How long would it take to produce 25.0 g of Cr from a solution of CrCl₃ by a current of 2.75A?

Given: Mass Cr = 25.0g; I = 2.75A

Find: Time

Cr⁺³ + 3e^- ® Cr_(s)

25.0 g Cr x (1 mole Cr/52.0 g Cr) x (3 mole e^-) = 1.44 mole e^-

1.44 mole e^- x (96,500C/1mole e^-) = 139,000C = q

139,000C x (1 AxS/1C) x (1/2.75A) = 50,500sec = 14 hours

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