20electrochemistry best slides 2

Chapter 20Electrochemistry

Dr. Peter [email protected]://www.chem.mun.ca/zcourses/1051.php

mailto:[email protected]

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Galvanic cells

Electrochemical cells fall into one of two basic types

Galvanic cells convert chemical energy into electrical energy (batteries)Electrolytic cells convert electrical energy into chemical energy.

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Here we put a piece of zinc

metal into a Cu2+ ion solution. A reaction occurs where we get Zn2+ ions and solid copper

deposited on the zinc surface.

4

This is an oxidation-reduction (redox)

process where electrons are

transferred from one chemical to another. One chemical loses

electrons in a process called

oxidation, while the other chemical gains

electrons in a process called

reduction.

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Zinc in Cu2+ solution is spontaneous

Since we actually see this reaction occurring, this

reaction must be spontaneous!

The reverse reaction, where we put copper metal into a Zn2+ ion

solution is non-spontaneous!

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

we can see that each zinc atom gives away 2 electrons

to a copper (II) ion to give us a copper atom and a zinc (II) ionin the spontaneous reaction!

Redox reaction

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To clarify the redox process, we often break a redox reaction down into two separate steps (half-reactions). In one half-reaction, a chemical loses electrons (is oxidized)

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

We call this the oxidation half-reaction.Notice that we are effectively treating electrons as a

“product” of the half-reaction.In the other half-reaction, we look at the reduction half-reaction, where a chemical gains electrons (is reduced)

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

Half-reactions

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The sum of these half-reactions must give us the overall reaction

of interest.

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

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

Zn (s) + Cu2+ (aq) + 2 e- Zn2+ (aq) + 2 e- + Cu (s)Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s)

Half-reactions

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Why do we call them “half-reactions”?Each half-reaction is written so we can see what is

happening to the electrons in the overall reaction.In reality a half reaction CANNOT occur by itself to

any great extent.The lost electrons in the oxidation half-reaction

MUST go somewhere.The gained electrons in the reduction half-reaction

MUST come from somewhere.Two half-reactions ALWAYS work together to give an overall reaction that can occur to a great extent.

Half-reactions

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Half-reactions

-n

reduction

oxidation

ne (aq) M (s) M

The electrons stay on the metal

electrode and are NEVER found in

solution!

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Since the chemicals are in direct contact with each other, the electron transfer occurs directly and we can’t use the electrons to do anything useful.

How can we separate the chemicals but allow the electrons to transfer indirectly so we can use them?

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Copper in Ag+ solution is spontaneous

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Since a half-reaction cannot take place by itself we need to connect the half-cells together. It turns out that we must make a circuit (two connections!) for the entire galvanic cell to work.

The half-reactions

to take place in separate

containers (called half-cells).

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The left half-cell has a solid copper electrode in a Cu2+ ion solution, while the right half-cell has a solid silver electrode in a Ag+ ion solution.

A wire can connect the two solid electrodes for the electrons to move through. To connect the two

solutions so that ions can move between the half-cells requires us to use a salt bridge, which is just

another solution of ions.

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Oxidation occurs at the copper

electrode, which we give the

special nameANODE

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

Since the anode collects the electrons that are lost, it has a negative charge and positive copper

ions leave the anode!

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Electrons move from the ANODE

to the silver electrode through

the wire

We can get them to do something useful, like light a

bulb!

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Reduction occurs at the silver

electrode, which we give the special

nameCATHODE

Ag+ (aq) + e- Ag (s)

Since the cathode collects the positive silver ions so they can gain the electrons, the cathode has a

positive charge!

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Positive ions leave the anode while the

cathode collects positive ions.Alternatively, negative ions

collect at the anode and move away

from the cathode

The ions are free to move through the salt bridge and are REQUIRED to complete the circuit!

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Overall, negative charges (electrons and

negative ions) are moving clockwiseOverall, positive

charges (positive ions and electron “holes”)

are moving counterclockwise

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The overall reaction is exactly the same as when we place solid copper in a Ag+ solution, but since we have separated the half-cells, we can look at the separate half-reactions

as they occur. Cu (s) Cu2+ (aq) + 2 e-

2 x [Ag+ (aq) + 1 e- Ag (s)]Cu (s) + 2 Ag+ (aq) + 2 e- Cu2+ (aq) + 2 e- + 2 Ag (s)

Cu (s) + 2 Ag+ (aq) Cu2+ (aq) + 2 Ag (s)

Note # of e-

must balance!

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Shorthand notation for galvanic cells

Drawing a diagram for a galvanic cell or describing it as we did in the previous problem is too time-

consuming to do on a regular basis.

We can use a shorthand notation!

Cu (s) + 2 Ag+ (aq) Cu2+ (aq) + 2 Ag (s)

Cu (s) | Cu2+ (aq) || Ag+ (aq) | Ag (s)

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Cu (s) | Cu2+ (aq) || Ag+ (aq) | Ag (s)

A single vertical line indicates a change in phase, like that between a solid

electrode and the solution its immersed in.

A double vertical line indicates a salt bridge.

What is not shown in the shorthand (but is always implied) is the wire connecting

the two electrodes to complete the circuit.

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Cu (s) | Cu2+ (aq) || Ag+ (aq) | Ag (s)

If we read the shorthand notation from left to right it says:

“A solid copper anode is in a solution of copper (II) ions which is connected by a salt bridge to a solution of silver

(I) ions into which a solid silver cathode has been placed. The

electrodes are connected by a wire.”

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We ALWAYS choose to write the cell notation with the oxidation reaction first and then the

reduction reaction.This means the leftmost chemical in the notation is ALWAYS the anode, while the rightmost chemical is ALWAYS the cathode. Additionally, the electrons ALWAYS flow from left to right through the wire, which is the way we read the shorthand.

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(through the wire connecting the

electrodes)

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Galvanic cells – Easy as ABC

Anode CathodeNegative Positive

Oxidation ReductionLeft Right

“The anode is the negative electrode where oxidation takes place. We put it on the left in

shorthand notation.”“The cathode is the positive electrode where

reduction takes place. We put it on the right in shorthand notation.”

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Other shorthand notation considerations

Sometimes gases are involved in galvanic cells.

Including them in the shorthand is easy once we realize the gas is just a separate phase and must be separated from other phases

by a vertical line.

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Consider this reaction

Cu (s) + Cl2 (g) Cu2+ (aq) + 2 Cl- (aq)Since we CAN’T use a gas as an electrode we need some solid substance to do that job. In this

case we bubble the gas by a carbon rodThe cell notation with the carbon acting as the cathode is

Cu (s) | Cu2+ (aq) || Cl2 (g) | Cl- (aq) | C (s)


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Cu (s) | Cu2+ (aq) || Cl2 (g) | Cl- (aq) | C (s)We can also be more specific by

giving concentration and pressure data for any of the aqueous or

gaseous chemicals of the system.e.g. Cu2+ (aq, 0.58 M) and Cl- (aq, 0.34 M) and Cl2 (g, 0.89 bar)


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Problem

Write the shorthand notation for a galvanic cell that uses the reaction

Fe (s) + Sn2+ (aq) Fe2+ (aq) + Sn (s)

Fe (s) | Fe2+ (aq) || Sn2+ (aq) | Sn (s)

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Problem

Write a balanced equation for the overall cell reaction and give a brief description of the galvanic cell represented by

Pb (s) | Pb2+ (aq) || Br2 (l) | Br- (aq) | Pt (s)Pb (s) + Br2 (l) Pb2+ (aq) + 2 Br- (aq)The reduction of Br2 to Br- occurs on

the surface of a Pt cathode

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Cell potentials for cell reactions

Electrons move from the copper anode through the wire to the silver cathode because it is energetically favourable for the electrons to move!An electron in a siver atom has less free energy

than the same electron in a copper atom. Much like a ball wants to roll down a hill so it ends

up where it has the lower potential energy, an electron wants to move to the atom where it has the

lower free energy.

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Potential

The difference in the free energy for the electrons in the anode and the cathode is somewhat like the slope from the top to the bottom of the hill.

If the hill is “steep”, the ball experiences more of the force of gravity than it does on a “gentle” hill.

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PotentialThe equivalent of the force of gravity to

the difference in the free energy of electrons in different atoms is called the

electromotive force (emf) – also known as the cell potential (E)

or the cell voltage (V). Like a ball on a steep hill, electrons are

under a “greater” force to transfer from the anode to the cathode when the cell potential has a larger magnitude.

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PotentialBecause there is a free energy difference for an electron in the anode as compared to the same electron in the cathode, the electron must lose free energy during the trip, just like a ball loses potential energy (as motion!) as it rolls down the hill. The free energy change is negative and so

the movement of the electrons is a spontaneous process!

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Potential

The free energy change is negative and so the movement of the electrons

is a spontaneous process!This occurs when the

potential is positive so a positive potential indicates a

spontaneous process

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Potential

We can get energy out of a ball (with its certain mass) rolling down a slope (the experienced gravity),

We can get energy out of an electron (with electrical charge)

that “rolls down the slope” that is the potential difference of electron

free energy between the two electrodes.

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PotentialIn terms of units, we can define

one Joule as the energy we get from a

charge of one Coulomb multiplied by the

potential of one volt.1 J = 1 C·V (one Coulomb-volt)

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Potential

A Coulomb is a very large unit of charge!

The charge on one electron is 1.60 x 10-19 C, so

one Coulomb is the charge of about

6 billion billion electrons!

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Potential

It is generally easier to talk about the charge of one mole of electrons, which we give

the special name of

Faraday Constant or faraday (F)

1 faraday = 6.022 x 1023 mol-1 e- x 1.60 x 10-19 C1 faraday = 9.65 x 104 C·mol-1

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PotentialWe can measure the potential between two electrodes with a voltmeter, which should give a positive reading when the positive terminal of the voltmeter is connected to

the positive electrode (the cathode), and

the negative terminal is connected to the negative electrode (the anode).

When the voltmeter gives a positive potential, we have identified the direction

of spontaneous change!

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Copper in Ag+ solution is spontaneous

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Standard cell potentialsCell potentials depend on many factors other than the chemicals in the system,

including the temperature, ion concentrations, and pressure.

Like in the thermodynamics chapter, where we defined a standard state of conditions for enthalpy tables, we can do the same to define

standard cell potentials E°.

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Standard cell potentialsWe can only measure a standard cell

potential if we have pure solids and liquids (activities of 1), all solution activities are 1 (@1 molL-1),

all gas activities are 1 (@1 bar), and the temperature is specified

(usually 25 °C).

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Standard cell potentials

Zn (s) | Zn2+ (aq) || Cu2+ (aq) | Cu (s)we can only measure the STANDARD cell potential if the [Zn2+] and [Cu2+] are both 1 molL-1 , and the copper and zinc electrodes are pure.

The E° for this cell is 1.10 V at 25 °C.

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Standard electrode potentials

The standard cell potential E for any galvanic cell can be expressed as the difference of the standard

electrode potentials for the cathode and the the anode E

cell = E(red),cathode - E

(red),anode

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The standard electrode potential depends on whether the electrode is

acting as the cathode or the anode.However, the process at the cathode (reduction) is the opposite process that would occur if it were happening

at the anode (oxidation).

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Reversing a process changes the sign of the electrode potential associated with the process.

Therefore we choose to report ALL standard electrode potentials

as reduction processesbecause for any specific electrode

E(red),cathode = - E(red),anode

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It would be nice to create a table of standard electrode potentials for all possible electrodes, then we could find standard cell potentials for any cell.

However, there is one problem! We’ve already seen that

half-reactions cannot occur without another half-reaction!


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We got around a problem like this in thermodynamics by defining the standard

enthalpy of formation of elements in their standard states as ZERO.

We can do the same for electrode potentials and set the potential for a

specific electrode as ZERO and measure all other electrode potentials in comparison to the standard.

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Standard hydrogen electrode

The standard electrode for potentials is the standard hydrogen electrode (S.H.E).The electrode consists of hydrogen gas at

1 bar bubbling through a 1 molL-1 (actually activity of 1) solution of H+ past

a platinum electrode. Therefore2 H+ (aq, a = 1) + 2 e- H2 (g, 1 bar)

E(red),cathode = EH+/H2 = 0 V

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Standard hydrogen electrode

If the oxidation reaction occurs instead in this half-cell as an anode, the overall reaction is

H2 (g, 1 bar) 2 H+ (aq, 1 molL-1) + 2 e-

Reversing a reaction changes the sign of the potential. For the S.H.E.

E(red),anode = - E(red),cathode = - EH+/H2

= 0 V

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The standard potential for this cell has been measured as 0.340 V at 25 C, and

our anode is the standard hydrogen

electrode!

V 0.340EEEEE2

2 HHCuCuanode(red),cathode(red),cell //

Pt (s) | H2 (g) | H+ (aq) || Cu2+ (aq) | Cu (s)

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We have defined the standard electrode potential of the

reduction of Cu2+ ions to solid Cu! This is also known as a

standard reduction potential.Cu2+ (aq) + 2 e- Cu (s)

E(red) = 0.340 V

V 0.340EEEEE2

2 HHCuCuanode(red),cathode(red),cell //

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If we reverse the half-reaction we’ll get the

standard OXIDATION potential for the oxidation of solid Cu to Cu2+

ions…Cu (s) Cu2+ (aq) + 2 e-

E(ox) = -0.340 V

V 0.340V 0.000 EE EECuCuanode(red),cathode(red),cell 2

/

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The standard potential for this cell has been measured as 0.763 V at 25 C, and our anode is the zinc

electrode!

V 0.763E -EE EE/ZnZn/HHanode(red),cathode(red),cell 2

2

Zn (s) | Zn2+ (aq) || H+ (aq) | H2 (g) | Pt (s)

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We have found the standard potential of the oxidation of solid

zinc to zinc ions! This is a

standard oxidation potential.Zn (s) Zn2+ (aq) + 2 e-

E(ox) = 0.763 V

V 0.763E-V 0.000EEE/ZnZnanode(red),cathode(red),cell 2

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If we reverse the half-reaction we’ll get the standard reduction

potential for Zn2+ ions to solid zincZn2+ (aq) + 2 e- Zn (s)

E(red) = -0.763 VWe report this value as the

standard electrode potential!

V 0.763E-V 0.000EEE/ZnZnanode(red),cathode(red),cell 2

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The standard cell potential for this cell can be calculated if we know the anode is the

zinc electrode and the cathode is the copper electrode!

V 1.103 V) (-0.763 - 0.340V

E-EEEE/ZnZn/CuCuanode(red),cathode(red),cell 22

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

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Standard electrode potentials E(red)

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Using standard electrode potentials

Using tabulated standard electrode potential data is accomplished much like a Hess’s Law problem with one very important difference!

Let’s consider

Zn (s) | Zn2+ (aq) || Ag+ (aq) | Ag (s)which has the balanced equation

2 Ag+ (aq) + Zn (s) 2 Ag (s) + Zn2+ (aq)

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Using standard electrode potentials

Oxidation Zn (s) Zn2+ (aq) + 2 e- Reduction 2 [Ag+ (aq) + 1 e- Ag (s)] 2 Ag+ (aq) + Zn (s) 2 Ag (s) + Zn2+ (aq)

Ecell = 1.563 VV 1.563 V) (-0.763 - 0.800V

E-EEEE/ZnZn/AgAganode(red),cathode(red),cell 2

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Like Hess’s Law, we look up the standard electrode potential reactions for both our sets

of chemicals and then reverse the half-reaction for the set undergoing oxidation while changing the sign of the electrode

potential (- E(red),anode!) .

However, we DO NOT multiply the potential for either half-reaction.

Why?

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PotentialRecall the potential is

like the slope of a hill. A hill does not change its slope if we have

two (or more!) balls rolling downhill instead of one ball.

Therefore the potential of an electrode does not change if we multiply to get the right

number of electrons!

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Problem

The standard cell potential for the following galvanic cell is 0.78 V

Al (s) | Al3+ (aq) || Cr2+ (aq) | Cr (s)The standard electrode potential for the Al electrode is -1.676 V. Calculate the standard electrode potential for the Cr electrode.

V 0.90- E

V 0.78 V) (-1.676 - EE-E

EEE

/CrCr

/CrCr/AlAl/CrCr

anode(red),cathode(red),cell

2

232

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Problem

Use the data from Table 20.1 to determine the Ecell for the redox reaction in which Fe2+ (aq) is oxidized to Fe3+ (aq) by MnO4

- (aq) in acidic solution. Also provide the overall reaction.Answer: Ecell = 0.74 V

5 Fe2+ (aq) + MnO4- (aq) + 8 H+ (aq)

5 Fe3+ (aq) + Mn2+ (aq) + 4 H2O (l)

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Free energy and electrical work

We’ve already seen for any system the energy free to do work is given by

G = H - TSat standard conditions, or

G = H – TSat non-standard conditions.

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Not all work has to be expansion (PV) work.

There are other types of work, one of which is electrical work!

There must be a connection between G and electrical work

done by a galvanic cell.

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We’ve seen for a spontaneous process that G < 0.

We’ve also seen for a galvanic cell the overall reaction is spontaneous, and the cell

potential is positive to indicate spontaneity.

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Therefore for a spontaneous process in a galvanic cell, the change in free energy (the

electrical work done) must be directly proportional to the negative of the potential.

G -Ecellor

G = welec = -kelec Ecell

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The constant of proportionality kelec must depend on two things. First, it depends on

how many electrons we have moved through the wire. Twice as many electrons should

mean twice as much work…

We will usually measure numbers of electrons in

moles and symbolize it by n.

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The constant of proportionality kelec must also depend on the charge of each

electron moving through the wire. Since we are already talking about moles of electrons, we should talk about the charge of one mole

of electrons.

We’ve already seen that the faraday (F) = 9.65 x 104 C·mol-1

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G = welec = -kelec Ecell

G = welec = -nFEcell

and at standard conditions


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Do the units match those for work?


= -(mol)(Cmol-1)(V)= CV = J

Yes! The units match those for work.

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Problem

Use the given electrode potential data to determine G for the reaction2 Al (s) + 3 Br2 (l)

2 Al3+ (aq, 1 M) + 6 Br- (aq, 1 M)

Al3+ (aq) + 3 e- Al (s) EAl3+/Al = -1.676 V

Br2 (l) + 2 e- 2 Br- (aq) EBr2/Br- = 1.065 VAnswer: Since Ecell is 2.741V and the rxn involves 6 mol of e-, then G is -1587 kJ.

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Problem

The hydrogen-oxygen fuel cell is a galvanic cell with a reaction

2 H2 (g) + O2 (g) 2 H2O (l)

Using the given data calculate Ecell for this reaction:

Gf (H2O) = -237.1 kJmol-1

Gf (H2) = 0.00 kJmol-1

Gf (O2) = 0.00 kJmol-1Answer: Since G is -474.2 kJ and the rxn involves 4 mol of e-, then Ecell is 1.229 V .

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Spontaneous change in redox reactions

We’ve already related the free energy change to the cell potential

G = -nFEcell

and we also know a spontaneous process

MUST HAVE G < 0

WHICH MEANS Ecell > 0 for ALL spontaneous electrochemical

(oxidation-reduction) processes.

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Problem

When sodium metal is added to seawater, which has [Mg2+] = 0.0512 M, no magnesium metal is obtained. According to the data below, should this reaction occur? What reaction does occur?Na+ (aq) + 1 e- Na (s) ENa+/Na = -2.713 V

Mg2+ (aq) + 2 e- Mg (s) EMg2+/Mg = -2.356 V

2 H2O (l) + 2 e- H2 (g) + 2 OH- (aq) EH2O/H2 = -0.828 V

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Problem answerFor the reaction

2 Na (s) + Mg2+ (aq) 2 Na+ (aq) + Mg (s)Ecell = 0.357 V and the reaction should be spontaneous. However, the reaction of sodium with water is 2 Na (s) + 2 H2O (l) 2 Na+ (aq) + H2 (g) + 2 OH- (aq)and has Ecell = 1.885 V and this reaction should also be spontaneous. Since this reaction is “more spontaneous” (higher Ecell) sodium preferentially reacts with water and not magnesium ions!

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Problem

Without using the data for a detailed calculation, explain why Sn2+ solutions must be protected from oxygen. One way to protect them is to add metallic (solid) tin.Sn4+ (aq) + 2 e- Sn2+ (aq) ESn4+/ Sn2+ = 0.154 V

Sn2+ (aq) + 2 e- Sn (s) ESn2+/ Sn = -0.137 V

O2 (g) + 4 H+ (aq) + 4 e- 2 H2O (l) EO2/H2O = 1.229 V

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Problem answer

For both possible reactions the reduction of oxygen is the cathode

half-cell reaction.

Since Ecell = E(red),cathode - E(red),anode, then the anode half-cell reaction that is

more negative will give the higher (“more spontaneous”) Ecell reaction that will

preferentially occur.

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Metals and acids

Some metals will react with acidic solutions to form H2 gas and metal ions in solution while others will not. We now know that those metals that do react with acid do so because the reaction is spontaneous while those that do not react do not because the reaction is non-spontaneous.

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Metals and acids

In MOST cases the reduction reaction of metals in acidic solutions is

2 H+ (aq) + 2 e- H2 (g)

E(red),cathode = EH+/H2 = 0 V

This is the standard hydrogen electrode half-reaction !

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Metals and acids

IF this is the preferred reduction (cathode) reaction, and since a spontaneous process

must have a positive potential then for a metal to react with the H+ of an acid means

Ecell = E(red),cathode - E(red),anode > 0(0 V) - E(red),anode > 0

E(red),anode < 0

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Metals and acids

The metals that CAN REACT with H+ are the ones with a negative Ered value like

Na (ENa+/Na = -2.713 V) or Al (EAl3+/Al = -1.676 V)

or Pb (EPb2+/Pb = -0.125 V)

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Metals and acids

The metals that CAN NOT REACT with H+ have a positive

Ered value like

Ag (EAg+/Ag = +0.800 V) or Au (EAu3+/Au = +1.52 V)

or Cu (ECu2+/Cu = +0.340 V)

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Metals and acids

Some acids, like HNO3 have a different preferred reduction

(cathode) reaction. For exampleNO3

- (aq) + 4 H+ (aq) + 3 e- NO (g) + 2 H2O (l)

E(red),cathode = ENO3-/NO

= +0.956 V

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Metals and acids

IF this is the preferred reduction (cathode) reaction, and since a spontaneous process

must have a positive potential then for a metal to react with the NO3

- of nitric acid means

Ecell = E(red),cathode - E(red),anode > 0(0.956 V) - E(red),anode > 0

E(red),anode < 0.956 V

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Metals and acids

In nitric acid all the metals that usually react with acids

will still react, but now Ag (EAg+/Ag = +0.800 V) WILL ALSO REACT!

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Ecell and Keq

We have two equations relating free energy to Keq and Ecell

G = -RT ln Keq

and

G = -nFEcell

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Ecell and Keq

By setting the equations equal to each other we find the relationship between cell potential and the thermodynamic equilibrium constant

-RT ln Keq = -nFEcell

Ecell = (RT/nF) ln Keq

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Ecell and Keq

Ecell = (RT/nF) ln KeqWe have two constants (R and F)

in this equation and we often perform reactions at 298.15 K.

With these three fixed values we can simplify this equation (but we

DON’T HAVE TO!)

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Ecell and Keq

Ecell = (0.025693 V/n) ln Keq

using R = 8.3145 JK-1mol-1

BE CAREFUL!This form ONLY applies

at 298.15 K!

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Everything is connected!

We COULD also put kinetics and rate constants and how they relate both to thermochemistry

and equilibrium in this diagram to show ALL the possible connections in the chemistry you’ve seen!

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Problem

Should the reaction of solid Al with Cu2+ ions go to completion at 25 C if Ecell for the reaction is 2.016 V?2 Al (s) + 3 Cu2+ (1 M) 3 Cu (s) + 2 Al3+ (1 M)

Answer: The reaction involves 6 moles of electrons, so Keq = e471 = very large, so the reaction goes to completion.

96

Problem

Should the reaction of solid Sn with Pb2+ ions go to completion at 25 C? Pb2+ (aq) + 2 e- Pb (s) EPb2+/Pb = -0.125 V

Sn2+ (aq) + 2 e- Sn (s) ESn2+/Sn = -0.137 V

Answer: Since Ecell = 0.012 V and 2 moles of electrons are involved in the process Keq = 2.5 and the reaction does not go to completion.

97

Ecell as a function of concentration

Zn (s) | Zn2+ (aq, 1M) || Cu2+ (aq, 1M) | Cu (s)We’ve seen that Ecell is 1.103 V for this reaction at standard conditions.

However, what happens to the cell at non-standard conditions?

Zn (s) | Zn2+ (aq, 0.10 M) || Cu2+ (aq, 2.0 M) | Cu (s)If we set up this cell and measure the potential, then Ecell is 1.142 V.

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Since the actual reaction isZn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s)then Le Chatalier’s Principle tells us

that decreasing [Zn2+] from 1 M to 0.10 M should shift the reaction towards

products and increasing [Cu2+] from 1 M to 2.0 M should shift the reaction

towards products as well.

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Both new concentrations serve to make the reaction more complete (or more

spontaneous), and so we expect a more positive potential!

Recall that

G = G + RT ln Qeq

andG = -nFEcell

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By substituting we see

-nFEcell = -nFEcell + RT ln Qeq

orEcell = Ecell - (RT/nF) ln Qeq

101


Sometimes we prefer log to ln!Since ln x = 2.3026 log x

we can change Ecell = Ecell - (RT/nF) ln Qeq

into the Nernst EquationEcell = Ecell - 2.3026

(RT/nF) log Qeq

102


For this reactionZn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s)

Qeq = aZn2+ / aCu2+ and so

Ecell = Ecell - 2.3026 (RT/nF) log{aZn2+/aCu2+}

103


If we plot Ecell versus log Qeq

we should get a straight line with a

slope of [- (2.3026 RT/nF)] and y-intercept of Ecell

104


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

Qeq = aZn2+ / aCu2+

105


Ecell = Ecell - (2.3026 RT/nF) log Qeq

Since R (8.3145 JK-1mol-1) and F (9.65 x 104 C·mol-1)

are constants, and if we choose the temperature to be 298.15 K, then we can

replace these three fixed values as we have done before (slide 93)

106

Nernst equation at 298.15 K

Ecell = Ecell – (0.0592 V/n) log Qeq

It makes the most sense to memorize the Nernst

Equation and substitute rather than remembering this

form for one temperature!

107

Problem

Calculate Ecell for the for the following galvanic cells at 298.15 K. Will the reactions be spontaneous?Al (s) | Al3+ (0.36 M)

|| Sn4+ (0.086 M), Sn2+ (0.54 M) | Pt(s)

Pt(s) | Cl2 (1 atm) | Cl- (1.0 M)

|| Pb2+ (0.050 M), H+ (0.10 M) | PbO2(s)

108

Problem data

Sn4+ (aq) + 2 e- Sn2+ (aq)ESn4+/Sn2+ = 0.154 V

Al3+ (aq) + 3 e- Al (s)EAl3+/Al = -1.676 V

PbO2 (s) + 4 H+ (aq) + 2 e- Pb2+ (aq) + 2 H2O (l)EPbO2/Pb2+ = 1.455 V

Cl2 (g) + 2 e- 2 Cl- (aq)ECl2/Cl- = 1.358 V

109

Problem answer3 Sn4+ (0.086 M) + 2 Al (s)

3 Sn2+ (0.54 M) + 2 Al3+ (0.36 M) Ecell = 1.830 V and Ecell = 1.815 V

PbO2 (s) + 4 H+ (0.10 M) + Cl2 (1 atm) Pb2+ (0.050 M) + 2 H2O (l) + 2 Cl- (1.0 M)

Ecell = 0.097 V and Ecell = 0.017 V Since both Ecell values are positive, both reactions

will be spontaneous at the given conditions.

110

Problem

For what ratio of [Sn2+] / [Pb2+] will the given cell reaction NOT be spontaneous in either direction?Sn (s) | Sn2+ (aq) || Pb2+ (aq) | Pb (s)

Pb2+ (aq) + 2 e- Pb (s) EPb2+/Pb = -0.125 V

Sn2+ (aq) + 2 e- Sn (s) ESn2+/Sn = -0.137 V

111

Problem answer

The reaction is NOT spontaneous in either direction ONLY at equilibrium, where Ecell = 0.

Since for this cell Ecell = 0.012 V the equilibrium occurs when

Qeq = [Sn2+] / [Pb2+] = Keq = 2.5

(see slide 96)

112

Concentration cells

We know if we mix two solutions of the same chemical but with different

concentrations, then the final solution will have a single uniform

concentration.

The mixing is a spontaneous process!

113

Concentration cells

We can set up the mixing process as an electrochemical cell!

The different concentrations in the two half-cellswill lead to a non-zero Ecell

114

Concentration cells

We can set up the mixing process as an electrochemical cell!

The different concentrations in thetwo half-cellswill lead to a

Ecell

different from

Ecell

115

Concentration cells for H+

Pt (s) | H2 (1 atm) | H+ (x M) || H+ (1 M) | H2 (1 atm) | Pt (s) H2 (g, 1 atm) 2 H+ (x M) + 2 e-

2 H+ (1 M) + 2 e- H2 (g, 1 atm)Net reaction: 2 H+ (1 M) 2 H+ (x M)

Ecell

will ALWAYSbe zero for a concentration

cell…

116


Net reaction: 2 H+ (1 M) 2 H+ (x M)Ecell = Ecell - 2.3026

(RT/nF) log Qeq

Ecell = 0 - 2.3026 (RT/nF) log x2/12

Ecell = 2.3026 (2RT/nF) (-log x)

117


Ecell = 2.3026 (2RT/nF) (-log x)if x = [H+] then (-log x) = pH

At 298.15 K we can replace our three fixed values R, F, and T to give

Ecell = 0.0592 V (pH)This is the basis for electronic

pH meters!

118

Concentration cells for finding Ksp

Ag (s) | Ag+ (sat’d AgI) || Ag+ (0.100 M) | Ag (s) Ag (s) 2 Ag+ (sat’d AgI) + e-

Ag+ (0.100 M) + e- Ag (s)Net reaction: Ag+ (0.100 M) Ag+ (sat’d AgI)

Ecell

will ALWAYSbe zero for a concentration

cell…

119


Net reaction: Ag+ (0.100 M) Ag+ (sat’d AgI)We measure Ecell for this concentration cell

and we find it to be 0.417 VSince

Q = [Ag+] / [Ag+]and

Ecell = Ecell – (0.0592 V/n) log Qeq

and n = 1

120


Net reaction: Ag+ (0.100 M) Ag+ (satd AgI)Ecell = Ecell - (0.0592 V) log Qeq

0.417 V = 0 - (0.0592 V) log Qeq

log Qeq = 0.417 V / (-0.0592 V)log Qeq = -7.044

Qeq = [Ag+]/(0.100 M) = 9.04 x 10-8

So [Ag+] = 9.04 x 10-9 M

121


Since the [Ag+] came from a saturated AgI solution, then

[Ag+] = [I-] = 9.04 x 10-9 Mand Ksp = [Ag+] [I-]

Ksp = (9.04 x 10-9) (9.04 x 10-9)Ksp = 8.3 x 10-17

122

Problem

If Ksp = 1.8 x 10-10 for silver chloride then what would be Ecell forAg (s) | Ag+ (sat’d AgCl) || Ag+ (0.100 M) | Ag (s)

Answer: Ecell = 0.23 V

123

Problem

Calculate the Ksp for lead iodide with the given concentration cell informationPb (s) | Pb2+ (sat’d PbI2) || Pb2+ (0.100 M) | Pb (s)

Ecell = 0.0567 VAnswer: Ksp = 7.1 x 10-9

124

Electrolysis and electrolytic cells

The reverse of every spontaneous chemical reaction is non-

spontaneous. If we apply electric current to a

chemical system, it is possible to force non-spontaneous chemical reactions

occur in a process called electrolysis, in what we call electrolytic cells.

125

Electrolysis and electrolytic cells

The potential we apply to the electrolytic cell must be greater than that for the spontaneous reaction, and must be applied

in the opposite direction.

Ebattery > -Ecell

126


Since we actually see this reaction occurring, this

reaction must be spontaneous!

The reverse reaction, where we put copper metal into a Zn2+ ion

solution is non-spontaneous!

127


If we want to get zinc from this cell, we must force the non-spontaneous

reaction to occur by applying a potential

in the direction opposite that for the

spontaneous process!

Reduction ALWAYS occurs at the

cathode!

Reduction! Oxidation!

Cathode is –ve! Anode is +ve!

128

Electrolysis as coupled reactions

When we are doing electrolysis we are using the spontaneous battery

reaction to drive the non-spontaneous electrolysis reaction.

battery reactants battery productsEbattery > 0 so G < 0

electrolysis reactants electrolysis productsEcell < 0 so G > 0

129

Electrolysis as coupled reactions

In our setup we are coupling (adding) the reactions which means we add

their potentials (or free energies).

battery reactants + electrolysis reactants electrolysis products + battery products

Esum = Ebattery + Ecell > 0 so Gsum < 0

130

Complicating factors in electrolysis

While adding the potentials to get the potential for the

coupled reaction is straightforward in theory,

in practice there are complicating factors!

131

Overpotentials

The electrolytic cell has electron transfers occurring at the surface of the electrodes. If

solutions are involved then there is generally a good contact to the electrode.

However, if gases are contacting the electrode the contact is problematic.

As the contact to the electrode gets worse we often need to apply an overpotential (extra

Eoverpotential) to make up for this problem.

132

Overpotentials

Ebattery > -(Ecell + Eoverpotential)For example, a solid platinum

electrode generally has a near-zero volt Eoverpotential

while the formation of H2 gas on the surface of a mercury cathode has

Eoverpotential 1.5 V

133

Competing reactions

If we set up an electrolytic cell expecting

Esum = Ebattery + Ecell > 0will give us the reaction we want we may be surprised when we get a completely

different reaction because

Ebattery + Eother > Ebattery + Ecell

134

Competing reactions

Ebattery + Eother > Ebattery + Ecell We saw in slides 78-79 that if we have competing reactions, then

the one that is “more spontaneous” will preferentially

occur.

135

Competing reactions

Often, but not always, when we do electrolysis in aqueous solutions we get

the competing reactions2 H2O (l) + 2e- H2 (g) + 2 OH- (aq)

at the cathode and2 H2O (l) O2 (g) + 4 H+ (aq) + 4 e-

at the anode.

136

Competing reactionsWhen we do electrolysis in aqueous

solution we must identify which of the two possible reduction reactions is “more spontaneous”

when forced andwe must identify which of the two possible oxidation reactions is “more spontaneous”

when forced.See pages 850-851 and Example Problem

20-11 in the text for more info on this very important topic

137

Non-standard conditions

Industrially we try to maximize product with minimum energy and money input. This often means that we do electrolysis

on cells at non-standard conditions, which means

Ecell Ecell

138

Electrodes

Platinum is an inert electrode that only provides a surface for the true reactants to transfer electrons.An active electrode is an actual reactant in the half-cell reaction.

Using a different electrode on one side of the electrolytic cell might change the

half-cell reaction on that side!

139

Quantitative aspects of electrolysis

140


If we pass 1 mole of electrons through the cell, from the balanced equation

Na+ (l) + e- Na (s)we see we will get 1 mole (23.0 g) of

solid sodium out. At the other electrode, where

2 Cl- (l) Cl2 (g) + 2 e-

we see that one mole of electrons is enough to give us one-half a mole (35.5

g) of Cl2.

141


How many electrons pass through the cell depends on the current, which is charge per unit time, (the ampere A, which is a C/s) and the time the current was allowed to pass though the cell…Charge (C) = Current (C/s) x time (s)

Charge (C) = Current (A) x time (s)

142


We saw earlier that one mole of electrons has a charge equal to one

Faraday1 F = 9.65 x 104 Cmol-1

moles of e- = Charge (C) / Faradaymoles of e- = (Current x time)

9.65 x 104 Cmol-1

143

The flowchart shows how to find the amount of substance that comes from electrolysis based on a known current and time.

If we want to know the current or time we used to get a certain

amount of substance, we reverse the order of the

flowchart.

144

ProblemHow many kilograms of aluminum can be produced in 8.00 h by passing a constant current of 1.00 x 105 A for an electrolytic cell with the following half reaction at the cathode?

Al3+ + 3 e- AlMolar mass of Al is 26.9815 gmol-1

Answer: 268 kg

145

Problem

A layer of silver is electroplated (an electrolytic process) on a coffee server using a constant current of 0.100 A. How much time is required to deposit 3.00 g of silver?

Molar mass of silver is 107.868 gmol-1

Answer: 7.45 hours

20electrochemistry best slides 2

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