Thermodynamics of Cells

8/2/2019 Thermodynamics of Cells

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1

The Gibbs Free Energy and Cell Voltage

When an amount of charge, Q, moves through a potential difference, ∆ E

w = - Q ∆ E

↑ b/c work done by the system

∆ E > 0 for galvanic (voltaic) cells

Recall, G = H – TS = E + PV – TS

For constant T and P : ∆G = ∆ E + P ∆V - T ∆S (usual case in electrochemical cells)

But ∆ E = q + w

Total work = P-V work + electrical work (w elect ) = - P ∆V + w elect

∆ E = q + w = q + w elect - P ∆V

∆G = q + w elect - P ∆V + P ∆V - T ∆S = = q + w elect - T ∆S

If the condition of reversibility is imposed upon the galvanic cells,q = q rev = T ∆S

Therefore, ∆G = q rev + w elect - T ∆S = T ∆S + w elect - T ∆S = w elect

And w elect = - Q ∆ E

If n moles of e − (nF coulombs of charge) pass through the external circuit of the

galvanic cell when it is operated reversibly and if ∆ E is the reversible voltage, then

∆G rxn = - Q ∆ E = - nF ∆ E (why ∆G rxn ? we’ll prove it in a bit!)

Electrical work is produced when ∆G rxn < 0

Standard States and Cell Voltages

∆G o

rxn = - n F ∆ E o, ∆ E

o(standard cell voltage): is the cell voltage in a galvanic cell in

which reactants and products are in their standard states. (Gases at 1 bar, solutes in 1 M,

metals in their pure stable states and at a specified temperature)

Example

A 6.0 Volt battery delivers a steady current of 1.25 A for a period of 1.50 hours.

Calculate the total charge in coulombs that passes through the circuit.

Q = It = (1.25 C s-1

)(1.5 hours)(3600 s hour -1

) = 6750 C

w elect = - Q ∆ E = - (6750 C)(6 V) = - 4.05 x 104

J (work done on the battery)

- (- 4.05 x 104

J) = + 4.05 x 104

J work done by the battery.



2

Half-Cell Potentials

• It is a very long and tedious job to tabulate all the conceivable galvanic cells

• We can avoid the previous job by tabulating the half-cell reduction potentials, E o

• These E o

values express the intrinsic tendency of a reduction half-reaction to

occur when the reactants and products are at standard states.

o

anode

o

cathode

o E E E −=∆

How E o’ s are measured?

Line notation of cell:

S.H.E. || Ag+

(aq,1M) | Ag( s)

Oxidation occurs in the SHE half-cell

and this electrode is the anode:

H2 ( g 1 bar) → 2H+

(aq,1M) + 2e −

2H+

(aq,1M) + 2e − → H2 ( g,1 bar)

E o

(H2 / H+, oxid. = E

o(H

+/ H2 , red.

≡ 0.00 V

Consider Cu2+

(aq,1 M)|Cu ( s) half-cell connected to SHE. We measureo

E ∆ = 0.34 V

Then we assign E o

for Cu2+

(aq,1 M) + 2e − → Cu ( s)

Now consider Zn2+

(aq,1 M)|Zn ( s) half-cell connected to SHE. We measureo E ∆ = - 0.76 V. Then we assign E

o= -.0.76V for Zn

2+(aq,1 M) + 2e − → Zn ( s)

Now we build the cell: Zn2+

(aq,1 M)|Zn ( s) || Cu2+

(aq,1 M)|Cu ( s) for which

V E E E o

anode

o

cathode

o 10.1)76.0(34.0 =−−=−=∆ (>0) - Cell is galvanic

Half-cell potentials are intensive properties, namely independent of the amount of the

reacting species.



3

1. All values are relative to SHE ( = reference electrode)2. Half-reactions are written as reductions (only reactants are oxidizing agents and

only products are the reducing agents)

3. The more positive the E o

the more readily the reaction occurs

4. Half-reactions are shown with b/c each can occur as reduction or oxidation

5. The half-cell that is listed higher at the table acts as the cathode

Writing spontaneous redox reactions

a) By convention, electrode potentials are written as reductions

b) When pairing two half-cells, you must reverse one reduction half-cell to produce anoxidation half-cell. Reverse the sign of the potential.

c) The reduction half-cell potential and the oxidation half-cell potential are added to

obtain the E ocell .

d) When writing a spontaneous redox reaction, the left side (reactants) must contain the

stronger oxidizing and reducing agents.

Example

Zn ( s) + Cu2+

(aq) → Zn2+

(aq) + Cu ( s) stronger stronger weaker weaker reducing oxidizing oxidizing reducing

agent agent agent agent



4

Example

Will Ag+

oxidize Zn ( s) or will Zn2+

oxidize Ag ( s)

Example

A voltaic cell houses the reaction between aqueous bromine and zinc metal:

Br 2+

(aq) + Zn ( s) → Zn2+

(aq) + 2Br -(aq) 83.1=o

cell E V

Calculateo

ebro E min giveno

zinc E = - 0.76 V.

The cell is voltaic 83.1=o

cell E V > 0 (reaction spontaneous) as written. Zn is being

oxidized and therefore it is the anode.

07.1)76.0(83.1 =−+=+=⇒−= o

anode

o

cell

o

cathode

o

anode

o

cathode

o

cell E E E E E E V =o

ebro E min

Example

A standard (Pt | MnO −4 , H+, Mn2+) half-cell consists of an inert electrode in contact with

a solution containing MnO−

4 (aq) , H+

(aq) and Mn2+

(aq) ions in standard states. Such a

half-reaction is assembled and connected to a standard Zn2+

(aq,1 M)|Zn ( s) half-cell/

a) Calculate ∆ E of the cell at 25oC.

In Appendix D we find:

(Rxn 1): MnO −

4 (aq) + 8H+

(aq) + 5e − → Mn2+

(aq) + 4H2O (l ) E o

= 1.507V

(Rxn 2): Zn2+ (aq) + 2e − → Zn ( s) E o = - 0.762V

Rxn 1 is the cathode because it lies above E o

(Zn2+

/Zn), So

269.2)762.0(507.1 +=−−=−=∆ o

anode

o

cathode

o E E E V

b) Write a balanced equation for the overall reaction

We multiply (Rxn 1) by 2 and (Rxn 2) by 5:

2MnO −

4 (aq) + 16H+

(aq) + 5Zn ( s) → 2Mn2+

(aq) + 5Zn2+

(aq) + 8H2O (l )

Number of e − involved: 10

Now check:

∆G o

rxn = [2∆G o

f (Mn2+

(aq)) +5∆G o

f (Zn2+

(aq))+ 8∆G o

f (H2O (l )) ] - [2∆G o

f (MnO −

4 (aq))

16∆G o

f (H+

(aq)) +5∆G o

f (Zn ( s)) ] = - 2194.5 kJ



5

V JC Cmol

Jmol x

F n

G E

e

o

rxno 27.227.2)485,96)(10(

105.2194 1

1

13

==−

=∆

=∆ −

−

−

Very good match!!!

Disproportionation

• A single chemical species is both oxidized and reduced

• This species must be able to give up electrons and accept electrons and in

addition the half-reaction in which it is reduced must lie higher in the table than

the half-reaction in which it is oxidized

• If this is the case then it drives the second reaction to go in reverse and

disproportionation occurs spontaneously

Example

(1): Cu+

(aq) + e−

→ Cu ( s) E o

= 0.521 V(2): Cu

2+(aq) + e − → Cu

+(aq) E

o= 0.153V

(1) is reduction and drives (2) as oxidation

2 Cu+

(aq) → Cu2+

(aq) + Cu ( s) E ocell = 0.521 – 0.153 = 0.368 V (>0)

Which means E orxn > 0 and ∆G o

rxn < 0 spontaneous. Yes it disproportionates.

Example

Decide whether Fe2+

(aq) in its standard state at 25oC is stable w/r/t

disproportionation.

In Appendix D we find:

(1): Fe3+

(aq) + e − → Fe2+

(aq) E o

= 0.771V

(2): Fe2+

(aq) + 2e − → Fe ( s) E o

= - 0.447V

In order for Fe2+

(aq) to disproportionate reaction (1) would have to be driven

backwards (oxidation): 3Fe2+

→ Fe ( s) + 2Fe3+

(aq)

And E o

= -0.447 – (+0.771) = -1.218 V (< 0) and ∆G o

rxn > 0 (non-spontaneous) and

therefore stable against disproportionation.

Oxygen as an oxidizing agent

O2 ( g ) + 4H+

(aq) + 4e − → 2H2O (l ) E o

= 1.229V good oxidizing agent!

O3 ( g ) + 2H+

(aq) + 2e − → O2 ( g ) + H2O (l ) E o

= 2.067V even better oxidizing

agent!



6

H2O2 (aq) + 2H+

(aq) + 2e − → 2H2O (l ) E o

= 1.776V (bleach + germicider)

O2 ( g ) +2H+

(aq) + 2e − → H2O2 (aq) E o

= 0.695V

H2O2 as a reducing agent can reduce only chemical species with reduction potentials

greater than 0.695 V.

Relative Reactivities of Metals

• Metals that can displace hydrogen from acid

Fe ( s) + 2H+

(aq) → H2 ( g ) + Fe2+

(aq) Why does this rxn go?

Fe ( s) → Fe2+

(aq) + 2e − E o

= - 0.44V and 2H+

(aq) + 2e − → H2 ( g ) E o

= 0.0 (SHE)

E ocell = 0.00 – ( - 0.44) = + 0.44 V (> 0) rxn will go!

• Metals that cannot displace hydrogen from acid

2Ag ( s) + 2H+

(aq) → 2Ag+

(aq) + H2 ( g ) E o

= 0.80V

2 x [Ag ( s) → Ag+

(aq) + e−

] and 2H+

(aq) + 2e−

→ H2 ( g ) E o

= 0.0 (SHE) E ocell = 0.00 – (0.80) = - 0.80 V (< 0) rxn will not go!

• Metals that can displace hydrogen from water

2H2O (l ) + 2e − → H2 ( g ) + 2OH − (aq) E o

= - 0.42 V

However, [OH − ] = 10-7

non-standard

Metals active enough to reduce H2O (l ) lie below ( – 0.42 V)

2Na ( s) + 2H2O (l )→

2Na

+

(aq) + H2 ( g ) + 2OH−

(aq)2 x [Na ( s) → Na

+(aq) + e − ] E

o= -2.71 V

2H2O (l ) + 2e − → H2 ( g ) + 2OH − (aq) E o

= -0.42V,

E ocell

= -0.42 – (-2.71) = 2.29 V Yes, Na can!

• Metals that displace other metals in solutes

Zn ( s) + Fe2+

(aq) → Zn2+

(aq) + Fe ( s)

Zn ( s) → Zn2+

(aq) + 2e − E o

= - 0.76 V

Fe2+

(aq) + 2e − → Fe ( s) E o

= - 0.44 V

E o

cell = -0.44 – (- 0.76) = 0.32 V

Concentration Effects and the Nernst Equation

Q RT GG o ln+∆=∆

∆G nFE o o= − ∆G nFE = −



7

Q RT GG o ln+∆=∆

− = − +nFE nFE RT Qo ln

Divide both sides – nF

−

−=

−

−+

−

nFE

nF

nFE

nF

RT

nF Q

o

ln

E E RT

nF Qo= − ln Nernst Equation (expresses the net driving force for a reaction)

E o: standard reduction potential

R: 8.314 J K -1

mol-1

= 8.314 V C K -1

mol-1

F : Faraday’s constant = 96,485 C mol-1

T : absolute temperature (K)

n: number of electrons in the half-reaction

Q: reaction quotient

Let us evaluate RT

nF at 25

oC (298.15K)

RT

nF

JK mol K

n Cmol n= =

− −

−

( . )( . )

( , )

.8314 29815

96 485

002571 1

1

And the Nernst Equation at 298.15K becomes, E E n

Qo= −00257.

ln

Recall that x x 10log303.2ln = and E E n

Qo= −0 05916

10

.log

Recall, at equilibrium ∆G = 0 ∴ ∆G RT K nFE RT K o o= − ⇒ − = −ln ln

And E RT

nF K o = ln

Let’s see in depth!

E E RT

nF Qo= − ln

When Q < 1 and thus [reactant] > [product], ln Q < 0, so E > E ocell

When Q = 1 and thus [reactant] = [product], ln Q = 0, so E = E ocell

When Q >1 and thus [reactant] < [product], ln Q > 0, so E < E ocell



8

The signs of ∆Goand E ocell determine the direction of the reaction at standard conditions.

Example

Suppose that the Zn |Zn2+

(aq) || MnO −

4 (aq) | Mn2+

(aq) |Pt cell is operated at pH 2.00

with [MnO −

4 ]= 0.12 M, [Mn2+

|] = 0.0010 M and [Zn2+

] = 0.015 M. Calculate the cell

voltage, ∆ E , at 25oC.

The reaction of this cell is:

2MnO −

4 (aq) + 16H+

(aq) + 5Zn ( s) → 2Mn2+

(aq) + 5Zn2+

(aq) + 8H2O (l ) , ∆ E o

= 2.27V

H MnO

Zn MnQ E E

o.2

)010.0()12.0(

)015.0()0010.0(ln

10

0257.027.2

][][

][][ln

10

0257.027.2ln

10

0257.0162

52

162

4

5222

=−=−=−∆=∆+−

++

Example

An electrochemical cell is set up at 25oC. One half-cell consists of a zinc anode

immersed in a 1.0 M solution of Zn(NO3)2. The second half-cell consists of a platinumcathode that has gaseous hydrogen bubbling over it at a pressure of 1.00 atm in a

solution of unknown hydrogen-ion concentration. The observed cell voltage is 0.472 V.

(a) Calculate the reaction quotient, Q

Anode: Zn ( s) →Zn2+

(aq) + 2e − E o

= - 0.762V

Cathode: 2H+

(aq) + 2e − → H2 ( g ) E o

= 0.00 V (SHE)



9

∆ E o

= 0.00 – (-0.762) = 0.762 V

The cell is not at the standard state

48.22)427.0762.0(0257.0

2)(lnln =−=∆−∆=⇒−∆=∆ E E

RT

nF QQ

nF

RT E E oo

948.22

108.5 xeQ ≈=

(b) Calculate the hydrogen-ion concentration in the second half-cell

M x H x H H

P ZnQ

H 59

22

2

103.1][108.5][

)00.1)(00.1(

][

][2 −+

++

+

=⇒===

Electrolytic Cells

In galvanic (voltaic) cells electrons are generated at the anode (-) and they are consumedst the cathode (+)

In an electrolytic cell, electrons come from an external power source, which supplies to

the cathode and removes them from the anode. Cathode (-), Anode (+)

The tin-copper reaction as the basis of a voltaic and an electrolytic cell

At standard conditions the

spontaneous reaction

between Sn and Cu2+

generates 0.48 V (voltaic)

If more than 0.48 V

supplied the same

apparatus and components

become an electrolytic

cell and the non-sponatneous reaction

between Cu and Sn2+

occurs.



10

Concentration Cells

In these cells the half-reactions are the same but the concentrations are different.

A concentration cell based on the Cu/Cu2+

half-reaction

The easiest is pH.

We construct a cell with cathode to be SHE and the unknown has the same apparatusdipping into an unknown [H

+] solution.

Oxidation: H2 ( g 1 atm) → 2H+

(aq, unknown) + 2e−

Reduction: 2H+

(aq, 1M) + 2e − → H2 ( g 1 atm)

Overall: 2H+

(aq, 1M) → 2H+

(aq, unknown)



11

]log[05916.0]log[2

05916.000.0

][

][log

2

05916.0 2

2

tan

2++

+

+

−=−=−= H H H

H E E

dard s

umknowno

cell cell

By measuring the cell potential we can get [H+] and consequently pH

pH Meter

Example

A concentration cell consists of two Ag/Ag+

half-cells. In half-cell A, electrode A dips

into 0.010 M AgNO3; in half-cell B, electrode B dips into 4.0 x 10-4

M AgNO3. What isthe cell potential at 298.15 K? Which electrode has a positive charge?

Ag+(aq, 0.010 M ) [half-cell A] → Ag

+(aq 4.0 x 10

-4 M ) [half-cell B]

V x

Ag

Ag E E

A

Bo

cell cell 0828.0010.0

1000.4log

1

05916.000.0

][

][log

1

05916.0 4

=−=−=−

+

+

Reduction occurs at the cathode electrode A, so electrode A has a positive charge.

Thermodynamics of Cells

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