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 orxn = - n F∆Eo, ∆Eo (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.
1
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, Eo
• These Eo values express the intrinsic tendency of a reduction half-reaction to
occur when the reactants and products are at standard states.
o
o
∆E o = Ecathode
− Eanode
How Eo’s are measured?
Line notation of cell:
S.H.E. || Cd2+ (aq A=1) | Cd(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)
Eo (H2 / H+, oxid. = Eo (H+ / H2 , red.
≡ 0.00 V
Consider Cu2+ (aq,1 M)|Cu (s) half-cell connected to SHE. We measure ∆E o = 0.34 V
Then we assign Eo for Cu2+ (aq,1 M) + 2e − → Cu (s)
Now consider Zn2+ (aq,1 M)|Zn (s) half-cell connected to SHE. We measure
∆E o = - 0.76 V. Then we assign Eo = -.0.76V for Zn2+ (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
o
o
∆E o = Ecathode
− Eanode
= 0.34 − ( −0.76) = 1.10V (>0) - Cell is galvanic
Half-cell potentials are intensive properties, namely independent of the amount of the
reacting species.
2
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 Eo 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
Example
Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s)
stronger
reducing
agent
stronger
oxidizing
agent
weaker
oxidizing
agent
weaker
reducing
agent
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:
o
= 1.83 V
Br2+ (aq) + Zn (s) → Zn2+ (aq) + 2Br - (aq) Ecell
o
o
Calculate Ebro
min e given E zinc = - 0.76 V.
o
= 1.83 V > 0 (reaction spontaneous) as written. Zn is being
The cell is voltaic Ecell
oxidized and therefore it is the anode.
o
o
o
o
o
o
o
Ecell
= Ecathode
− Eanode
⇒ Ecathode
= Ecell
+ Eanode
= 1.83 + ( −0.76) = 1.07 V = Ebro
min e
3
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 E we find:
(Rxn 1): MnO 4− (aq) + 8H+ (aq) + 5e − → Mn2+ (aq) + 4H2O (l) Eo = 1.507V
(Rxn 2): Zn2+ (aq) + 2e − → Zn (s) Eo = - 0.762V
Rxn 1 is the cathode because it lies above Eo (Zn2+/Zn), So
o
o
∆E o = Ecathode
− Eanode
= 1.507 − ( −0.762) = +2.269 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 orxn = [2∆G of (Mn2+ (aq)) +5∆G of (Zn2+ (aq))+ 8∆G of (H2O (l)) ] - [2∆G of (MnO 4− (aq))
16∆G of (H+ (aq)) +5∆G of (Zn (s)) ] = - 2194.5 kJ
o
∆Grxn
− 2194.5 x10 3 Jmol −1
∆E =
=
= 2.27 JC −1 = 2.27V Very good match!!!
−1
ne F
(10)(96,485Cmol )
o
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
disporoportionation occurs spontaneously
4
Example
(1): Cu+ (aq) + e − → Cu (s) Eo = 0.521 V
(2): Cu2+ (aq) + e − → Cu+ (aq) Eo = 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 orxn < 0 spontaneous. Yes it disproportionates.
Example
Decide whether Fe2+ (aq) in its standard state at 25oC is stable w/r/t
disproportionation.
In Appendix E we find:
(1): Fe3+ (aq) + e − → Fe2+ (aq) Eo = 0.771V
(2): Fe2+ (aq) + 2e − → Fe (s) Eo = - 0.447V
In order for Fe2+ (aq) to disproportionate reaction (1) would have to be driven
backwards (oxidation): 3Fe2+ → Fe (s) + 2Fe3+ (aq)
And Eo = -0.447 – (+0.771) = -1.218 V (< 0) and ∆G orxn > 0 (non-spontaneous) and
therefore stable against disproportionation.
Oxygen as an oxidizing agent
O2 (g) + 4H+ (aq) + 4e − → 2H2O (l) Eo = 1.229V good oxidizing agent!
O3 (g) + 2H+ (aq) + 2e − → O2 (g) + H2O (l) Eo = 2.067V even better oxidizing
agent!
H2O2 (aq) + 2H+ (aq) + 2e − → 2H2O (l) Eo = 1.776V (bleach + germicider)
O2 (g) +2H+ (aq) + 2e − → H2O2 (aq) Eo = 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 − Eo = - 0.44V and 2H+ (aq) + 2e − → H2 (g) Eo = 0.0 (SHE)
E ocell = 0.00 – ( - 0.44) = + 0.44 V (> 0) rxn will go!
5
• Metals that cannot displace hydrogen from acid
2Ag (s) + 2H+ (aq) → 2Ag+ (aq) + H2 (g) Eo = 0.80V
2 x [Ag (s) → Ag+ (aq) + e − ] and 2H+ (aq) + 2e − → H2 (g) Eo = 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) Eo = - 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 − ] Eo = -2.71 V
2H2O (l) + 2e − → H2 (g) + 2OH − (aq) Eo = -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 − Eo = - 0.76 V
Fe2+ (aq) + 2e − → Fe (s) Eo = - 0.44 V
E ocell = -0.44 – (- 0.76) = 0.32 V
Concentration Effects and the Nernst Equation
∆G = ∆G o + RT ln Q
∆G o = − nFE o
∆G = − nFE
∆G = ∆G o + RT ln Q
− nFE = − nFE o + RT ln Q
Divide both sides – nF
− nFE − nFE o
RT
ln Q
=
+
− nF
− nF
− nF
RT
E = Eo −
ln Q Nernst Equation (expresses the net driving force for a reaction)
nF
Eo: 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
6
Let us evaluate
RT
at 25oC (298.15K)
nF
RT (8.314 JK −1mol −1 )( 29815
. K ) 0.0257
=
=
nF
n(96,485Cmol −1 )
n
o
And the Nernst Equation at 298.15K becomes, E = E −
o
Recall that ln x = 2.303 log10 x and E = E −
0.0257
ln Q
n
0.05916
log10 Q
n
Recall, at equilibrium ∆G = 0 ∴ ∆G o = − RT ln K ⇒ − nFE o = − RT ln K
o
And E =
RT
ln K
nF
Let’s see in depth!
E = Eo −
RT
ln Q
nF
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
The signs of ∆Go and E ocell determine the direction of the reaction at standard conditions.
Example
7
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 25 oC.
The reaction of this cell is:
2MnO 4− (aq) + 16H+ (aq) + 5Zn (s) → 2Mn2+ (aq) + 5Zn2+ (aq) + 8H2O (l) , ∆Eo = 2.27V
0.0257
0.0257 [ Mn 2+ ]2 [ Zn 2+ ]5
0.0257 (0.0010) 2 (0.015) 5
∆E = ∆E −
ln Q = 2.27 −
ln
= 2.27 −
ln
= 2.14V
10
10
[ MnO4− ]2 [ H + ]16
10
(0.12) 2 (0.010)16
o
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 platinum
cathode 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 − Eo = - 0.762V
Cathode: 2H+ (aq) + 2e − → H2 (g) Eo = 0.00 V (SHE)
∆Eo = 0.00 – (-0.762) = 0.762 V
The cell is not at the standard state
∆E = ∆E o −
RT
nF
2
ln Q ⇒ ln Q =
( ∆E o − ∆E ) =
(0.762 − 0.427) = 22.48
nF
RT
0.0257
Q = e 22.48 ≈ 5.8 x10 9
(b) Calculate the hydrogen-ion concentration in the second half-cell
Q=
[ Zn 2+ ]PH 2
+ 2
[H ]
=
(1.00)(1.00)
= 5.8 x109 ⇒ [ H + ] = 1.3x10 −5 M
+ 2
[H ]
Electrolytic Cells
In galvanic (voltaic) cells electrons are generated at the anode (-) and they are consumed
st 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 (+)
8
The tin-copper reaction as the basis of a voltaic and an electrolytic cell
Concentration Cells
9
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 apparatus
dipping 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)
E cell = E
o
cell
2
[ H + ]umknown
0.05916
0.05916
−
log + 2
= 0.00 −
log[H + ]2 = −0.05916 log[H + ]
2
[ H ]s tan dard
2
By measuring the cell potential we can get [H+] and consequently pH
pH Meter
10
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 is
the 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]
Ecell = E
o
cell
0.05916
[ Ag + ]B
0.05916
4.00 x10 −4
−
log
= 0.00 −
log
= 0.0828V
1
[ Ag + ] A
1
0.010
Reduction occurs at the cathode electrode A, so electrode A has a positive charge.
11