The Gibbs Free Energy and Cell Voltage
When an amount of charge, Q, moves through a potential difference, ∆Ew = - 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 ∆Go
rxn = - n F∆E
o, ∆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.
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 anode o cathode o E E E = − ∆
How Eo’sare 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 ∆Eo= 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 o
E
∆ = - 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
V E E E o anode o cathode o = − =0.34−(−0.76) =1.10 ∆ (>0) - Cell is galvanic
Half-cell potentials are intensive properties, namely independent of the amount of the reacting species.
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 stronger weaker weaker
reducing oxidizing oxidizing reducing agent agent agent 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: Br2+ (aq) + Zn (s) → Zn2+ (aq) + 2Br - (aq) o =1.83 cell E V Calculate o e bro E min given o zinc E = - 0.76 V. The cell is voltaic o =1.83
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 e bro 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 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
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) + 8H 2O (l) Number of e−involved: 10 Now check: ∆Go rxn = [2∆Gof (Mn 2+ (aq)) +5∆Go f (Zn 2+ (aq))+ 8∆Go f(H2O (l)) ] - [2∆Gof(MnO−4(aq)) 16∆Go f (H + (aq)) +5∆Go f (Zn (s)) ] = - 2194.5 kJ V JC Cmol Jmol x F n G E e o rxn o 2.27 2.27 ) 485 , 96 )( 10 ( 10 5 . 2194 1 1 1 3 = = − = ∆ = ∆ − − −
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 disporoportionation occurs spontaneously
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) Eo
cell = 0.521 – 0.153 = 0.368 V (>0) Which means Eo
rxn > 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 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 ∆Go
rxn > 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−→ H
2 (g) Eo = 0.0 (SHE) Eo
• 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−→ H
2 (g) Eo = 0.0 (SHE) Eo
cell = 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, Eo
cell = -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 Eo
cell = -0.44 – (- 0.76) = 0.32 V
Concentration Effects and the Nernst Equation
Q RT G G= ∆ o + ln ∆ ∆Go = −nFEo ∆G= −nFE Q RT G G= ∆ o + ln ∆ −nFE= −nFEo +RTlnQ
Divide both sides – nF
− − = − − +− nFE nF nFE nF RT nF Q o ln E E RT nF Q o
= − ln Nernst Equation (
expresses the net driving force for a reaction)
Eo: standard reduction potentialR: 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 0 0257 1 1 1
And the Nernst Equation at 298.15K becomes, E E
n Q
o
= − 0 0257. ln
Recall that lnx =2.303log10 x and E E
n Q
o
= − 0 05916. log10
Recall, at equilibrium ∆G = 0 ∴ ∆Go = −RTlnK ⇒ −nFEo = −RTlnK And E RT
nF K
o = ln
Let’s see in depth!
E E RT
nF Q o
= − ln
When Q < 1 and thus [reactant] > [product], ln Q < 0, so E > Eo cell When Q = 1 and thus [reactant] = [product], ln Q = 0, so E = Eo
cell When Q >1 and thus [reactant] < [product], ln Q > 0, so E < Eo
cell
The signs of ∆Go and Eo
cell determine the direction of the reaction at standard conditions. Example
Suppose that the Zn |Zn2+(aq) || MnO−
4(aq) | Mn
2+(aq) |Pt cell is operated at pH 2.00 with [MnO−
4]= 0.12 M, [Mn
2+|] = 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) + 8H 2O (l) , ∆Eo = 2.27V V H MnO Zn Mn Q E E o 2.14 ) 010 . 0 ( ) 12 . 0 ( ) 015 . 0 ( ) 0010 . 0 ( ln 10 0257 . 0 27 . 2 ] [ ] [ ] [ ] [ ln 10 0257 . 0 27 . 2 ln 10 0257 . 0 16 2 5 2 16 2 4 5 2 2 2 = − = − = − ∆ = ∆ +− ++ 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
48 . 22 ) 427 . 0 762 . 0 ( 0257 . 0 2 ) ( ln ln ⇒ = ∆ −∆ = − = − ∆ = ∆ E E RT nF Q Q nF RT E E o o 9 48 . 22 5.8x10 e Q = ≈
(b) Calculate the hydrogen-ion concentration in the second half-cell M x H x H H P Zn Q H 9 5 2 2 2 10 3 . 1 ] [ 10 8 . 5 ] [ ) 00 . 1 )( 00 . 1 ( ] [ ] [ 2 + − + + + = ⇒ = = = 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 (+)
The tin-copper reaction as the basis of a voltaic and an electrolytic cell
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−→ H 2 (g 1 atm) Overall: 2H+ (aq, 1M) → 2H+ (aq, unknown) ] log[ 05916 . 0 ] log[ 2 05916 . 0 00 . 0 ] [ ] [ log 2 05916 . 0 2 2 tan 2 + + + + − = − = − = H H H H E E dard s umknown o cell cell
By measuring the cell potential we can get [H+] and consequently pH
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 10-4M 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 10-4M) [half-cell B] V x Ag Ag E E A B o cell cell 0.0828 010 . 0 10 00 . 4 log 1 05916 . 0 00 . 0 ] [ ] [ log 1 05916 . 0 4 = − = − = ++ −