Galvanic Cells - Edwardsville School District 7 - Edwardsville

Galvanic Cells

Chapter 17 Sections 1, 2 &4

Electrochemistry e- e-

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Electrochemistry- Standard Reduction Potentials e- e-

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Activity Series

Metals

Halogens

Li

Rb

K

Ba

Ca

Na

Mg

Al

Mn

Zn

Cr

Fe

Ni

Sn

Pb

H

Cu

Hg

Ag

Pt

Au

F

Cl

Br

I


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Electrochemistry – Galvanic Cells

Electrons are transferred directly when reactants collide

No work is obtained – instead heat is released

How can we obtain work?

Separate the oxidizing and reducing agents

require the e- to go through a wire

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MnO41- + 8H+ + 5e- Mn2+ + 4H2O

Fe2+ Fe3+ + 1e-

Reduction - Mn

Oxidation - Fe

Oxidizing agent - MnO41-

Reducing Agent – Fe2+

Current flows for a second then stops

Something more needs to added.

p. 792


It needs a salt bridge

Solutions must be connected so that ions can flow and keep the net

charge in each container at zero.

A salt bridge is a U-tube with electrolyte (pastey stuff) or porous disk.

Choose a substance that would be noninteractive (something made of

spectator ions if it’s a paste)

It allows ions to flow without mixing the solutions

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A galvanic cell changes chemical energy to electrical energy.

Components

1. Two separate solutions (oxidizing & reducing agent)

2. A wire

3. A salt bridge

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REDUCING

AGENT

OXIDIZING

AGENT

ANODE

oxidation

X Y + e-

p. 793

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CATHODE

reduction

X + e- Y


Be able to diagram a cell, label the parts and the flow.

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ZnSO4

REDUCING

AGENT

CuSO4

OXIDIZING

AGENT

ANODE

oxidation

Zn2+ Zn + e-

p. 793

e-

CATHODE

reduction

Cu + 2e- 2Cu2+

Zn Cu

Zn2+

SO4 2- Cu2+

SO4 2-

What is the

cathode?

Reduction

occurs at the

cathode.

Copper has a

greater

reduction

potential

Which way

do e- flow?

Anode to Cathode

What is

happening at

the salt bridge?

cations

anions

What are the

agents?

standard

reduction

potentials

see

the

atoms

Electrochemistry- Standard Reduction Potentials

Line Notation for the cell – so you don’t have to draw the cell

• Anode on the left; Cathode on the right

• Separate the half cells with a ||

• Separate the electrode from the solution with a |

ANODE CATHODE

Zn(s) | Zn 2+(aq) || Cu3+

(aq) | Cu(s)

Oxidation chamber || Reduction chamber

electrode | soln || soln | electrode

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Cell Potential – the “pull” or driving force that makes electrons go from the reducing agent to the oxidizing agent

•The potential to lose e- The push from the element mostly likely to loose to the one that will gain

• E cell

•Also called electromotive force, emf, of the cell

•Unit = volt = 1 joule of work per columb of charge transferred

– V = 1 J / 1 C

•Measured with a voltmeter

– Would measure less than cell potential

– Because it doesn’t measure the frictional heating of the wire

– New voltmeters use a negligible amount of current so they are used

•Measured with a potentiometer

– Variable voltage device (powered from a cell circuit)

– Adjusted so no current flows in the cell circuit

– Then cell potential = voltage setting but opposite sign

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Water only spontaneously flows one way in a waterfall.

Likewise, electrons only spontaneously flow one way in a redox reaction—from higher to lower potential energy.

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If you don’t want the electrode to participate in the reaction pick a “chemically inert” metal or element.

Like Au or Pt. But what would be cheaper?

Carbon. Really? Does carbon conduct electricity?

Graphite (gr) does!

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How would I know how any metal compares against any other?

We need a standard – one that can be oxidized or reduced.

2H+ + 2e- H20 reduction

H20 2H+ + 2e- oxidation

H+ H2+ possible but not common

The standard’s E cell would be zero.

Standard Hydrogen Electrode or SHE half cell

H2 can’t be solid so use a Pt electrode

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Can measure total potential of the cell E cell = 0.76 V

Can’t measure the potential of the half reactions ( or half cells)

Setting the standard potential for the hydrogen half reaction to zero

Allows us to assign values to all other half reactions

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2H+ + 2e- H2 Zn Zn2+ + 2e-

p. 794


E cell = E 2H+ H2 + E Zn Zn2+

0.76 = 0.000V + x

E Zn Zn2+ = 0.76 V

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2H+ + 2e- H2 Zn Zn2+ + 2e-

Denotes “standard state”

not the same as STP

Standard conditions p. 246

Compounds

gases = 1 atm

soln = 1M

Liquid or solid = pure

Element

1 atm

25 C

p. 794

Oxidation Reduction Reactions

Oxidation States

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E cell = E Zn Zn2+ + E Cu2+ Cu

1.10V = 0.76 V + E Cu2+ Cu

E Cu2+ Cu = 0.34 V

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Cu2+ + 2e- Cu Zn Zn2+ + 2e-

p. 795


The half reaction with the largest potential will run as a reduction (as written)

The other will be oxidation; so it will run in reverse

So…

F2 + 2e- 2F1- E = 2.87 V (reduction)

but

2F1- F2 + 2e- E = - 2.87 V (oxidation)

E cell = E cathode + E anode reverse the sign of the anode

E cell = E cathode - E anode

Multiplying the half reaction so that the e- are equal doesn’t

change the E

E is an intensive property

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About Standard Reduction Potentials

•A negative reduction potential means it will most likely oxidize

•If the E cell is greater than zero you will get current,

spontaneously.

•If the E cell is zero then the system has reached equilibrium.

The cell is “dead”.

•Values in the table are predicting perfect conditions. You probably won’t get this in the lab. Why?

– Voltmeter

•

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This is the chart given with the AP Exam.

Half reactions are written as reductions


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The strongest oxidizers

(oxidizing agents) have the

most positive reduction

potentials.

The strongest reducers

(reducing agents) have the

most negative reduction

potentials.


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The greater the difference

between the two, the

greater the voltage of the

cell.


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Will Br2 oxidize H2O2?

yes

Will Cd reduce Ag+?

yes

Can Al oxidize Au?

What are common metals in nature?

What are common metal ions in nature?

What are common non metal ions in nature?

a. Consider a galvanic cell based on the reaction

Al3+ (aq)

+ Mg (s) Al (s) + Mg2+(aq)

Find the half reactions, balance the cell reaction and calculate E for the cell.

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Electrochemistry- Sample Exercise 17.1 Page 797

b. Consider a galvanic cell based on the reaction

MnO4 1-

(aq) + H+ (aq) + ClO3

1- (aq) ClO4

1- (aq) + Mn2+

(aq) + H2O (l)

Find the half reactions, balance the cell reaction and calculate E for the cell.

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Electrochemistry- Sample Exercise 17.1 Page 797

What if you have a substance that is not solid for the electrode?

Look at the example on page 799.

It is the same reaction as our Redox Titration Lab Fe2+ + MnO4 1-

In the example iron is solid but MnO4 1- / Mn2+ is a solution.

Fe2+ Fe

MnO4 1- Mn2+

What is the line notation?

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Fe ǀ Fe2+ ǁ H+, MnO4 1-, Mn2+ ǀ Pt

Include the inert

electrode

Don’t include

the water

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Electrochemistry- Try #29 Page 831

© 2009, Prentice-Hall, Inc.

Electrochemistry- Gold/Nickel Voltaic Cell e- e-

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Au Ni

Au3+(aq)

X-(aq) X-(aq)

Ni2+(aq)

K+(aq) NO3-(aq)

Au3+ + 3e- Au

E0 = +1.50 V Ni2+ + 2e- Ni E0 = -0.25 V

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A cell will always run spontaneously in the direction that produces a positive cell

potential

Nickel is the anode E cell = E cathode - E anode flip the nickel

Current flows from anode to cathode

Au3+ + 3e- Au

E0 = 1.50 V

reduction: cathode

Ni Ni2+ + 2e-

E0 = 0.25 V

oxidation: anode

Au Ni

Au3+(aq)

X-(aq) X-(aq)

Ni2+(aq)

K+(aq) NO3-(aq)

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E0(V)

Au3+ + 3e- Au 1.50

Ni Ni2+ + 2e- 0.25

2 Au3+ + 3 Ni 2 Au + 3 Ni2+ 1.75


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Line Notation:

ANODE CATHODE

Ni(s)|Ni2+ (aq, 1 M)||Au3+ (aq, 1 M)|Au(s)

oxidation reduction


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© 2009, Prentice-Hall, Inc.

Electrochemistry- Aluminum/Nickel Cell e- e-

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Al Ni

Al3+(aq)

X-(aq) X-(aq)

Ni2+(aq)

K+(aq) NO3-(aq)

Al3+ + 3e- Al

E0 = -1.66 V Ni2+ + 2e- Ni E0 = -0.25 V

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E0(V)

Al Al3+ + 3e- 1.66

Ni2+ + 2e- Ni -0.25

2Al + 3Ni 2+ 3Ni + 2Al3+ 1.41


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Line Notation:

ANODE CATHODE

Al(s) | Al3+(aq, 1 M) || Ni2+

(aq, 1 M) | Ni (s)

oxidation reduction


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Sketch the Ag/Ag+ & Zn/Zn2+ galvanic cell. Show the direction of e- flow and the direction of ion migration. Calculate E0 for the cell. Give the line notation for the cell.

Electrochemistry- Extra Problem e- e-

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Zn Ag

Zn2+(aq)

X-(aq) X-(aq)

Ag+(aq)

K+(aq) NO3-(aq)

Zn2+ + 2e- Zn

E0 = - 0.76 V

Ag+ + e- Ag

E0 = + 0.80 V

e- ?

Sketch a system having a solid lead electrode and a lead(IV) oxide electrode immersed in sulfuric acid. Show the direction of e- flow and the direction of ion migration. Calculate E0 for the cell. Give the line notation for the cell.

Electrochemistry- Extra Problem e- e-

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Pb

Pb

O2

e- ?

H2SO4(aq)

PbSO4 + 2e- Pb + SO42-

E0 = - 0.35 V

PbO2 + 4H+ + SO4

2- + 2e- PbSO4 + 2H2O

E0 = + 1.69V

The Lead-Acid Car Battery


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Cu(s) + 2Ce 4+ (aq)

2Cu 2+ (aq)

+ Ce 3+ (aq)

Remember E is positive. Treat this like an equilibrium.

Increase [Ce 4+ ], which direction is favored?

Forward. Increase the driving force on e-, increase the E Increase [Cu 2+ ] or [Ce 3+], which direction is favored?

Reverse. Decrease the driving force on e-, decrease the E

Sample Exercise 17.5 page 803

2Al + 3 Mn2+ 2Al3+ + 3 Mn

a. [Al3+] = 2.0M; [Mn2+] = 1.0M

b. [Al3+] = 1.0M; [Mn2+] = 3.0M

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Electrochemistry- Concentration on E

Al3+ decrease E < 0.48V

Mn2+ increase E > 0.48V

Cell Potentials depend on concentration

Can construct cells with same components but different concentrations

Ag ǀ 0.10M Ag2+ ǁ 1.0 M Ag2+ ǀ Ag

E cell =0.80V – 0.80 V = 0

However since the concentration is not equal,

the half cell potentials are not 0.80V.

This is a concentration cell. Voltages are typically small

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The Nernst Equation

Derived from the dependence of free energy on concentration

*** Will learn about this after thermodynamics***

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n E = E cell -

0.0591 log (Q)

moles of e-

Like Keq but not necessarily at equilibrium

Remember (l) and (s) don’t go into the Keq

Why not? It’s all about concentration – solids and liquids have a definite volume

•The M for solids and liquids is 1 M= mol/L

•If you change moles for solids and liquids then the volume changes proportionally

Quotient = Keq

[P]x

[R]y

The Nernst Equation

Derived from the dependence of free energy on concentration

*** Will learn about this after thermodynamics***

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n E = E cell -

0.0591 log (Q)

moles of e-

Trying to get the max voltage?

•Choose the half reactions that give max volts – be practical Li or F won’t work

•Make the reactant concentration greater than 1 & the product concentration less than 1

That way [P]x / [R]y is 0.XX and then log 0.XX is a negative

number and you’re adding to the E cell

Quotient = Keq

[P]x

[R]y

Handy ideas about logs

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Electrochemistry- Sample Exercise 17.7 page 806

= log 1 – log x = -log x log 1

x log x 2 = 2 log x

log 100 = 2

Sig. Figs for Logarithms For any log, the number to the left of the decimal point is called the characteristic, and the number to the right of the decimal point is called the mantissa. The characteristic only locates the decimal point of the number, so it is usually not included when determining the number of significant figures. The mantissa has as many significant figures as the number whose log was found. Example 1: log 5.43 x 1010 = 10.735 The number has 3 significant figures, but its log ends up with 5 significant figures, since the mantissa has 3 and the characteristic has 2.

log xy = log x + log y

log x y = y log x log y√ x = log x1/y = (1/y ) log x

Describe the cell based on the following half reactions:

VO2+

(aq) + 2H1+ (aq) + e- VO 2+

(aq) + H2O (l)

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

T = 25 C

[VO2+] = 2.0 M

[H1+] = 0.50M

[VO 2+ ] = 1.0 x 10-2 M

[Zn2+ ] = 1.0 x 10-1 M

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Describe the cell based on the following half reactions:

VO2+

(aq) + 2H1+ (aq) + e- VO 2+

(aq) + H2O (l)

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

Where T= 25 C

[VO2+] = 2.0 M

[H1+] = 0.50M

[VO 2+ ] = 1.0 x 10-2 M

[Zn2+ ] = 1.0 x 10-1 M

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Solution

1. Balance First

2. Calculate the standard cell potential

3. Use the Nernst Equation

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Galvanic Cells - Edwardsville School District 7 - Edwardsville