Electrochemistry and Thermodynamics
Voltaic cells can do work because ∆G for the redox reaction is negative. The voltaic cell continues to do work until
an equilibrium is reached between the reactants and products.
As the cell does work, electrons flow, the redox reaction takes place, the concentrations of reactants and
products change towards equilibrium, and the cell potential continually decreases.
“Think about it”: At equilibrium the voltaic cell has:
E = ?
Q = ?
∆G = ?
REMEMBER: emf, E° or E, are INTENSIVE properties (J/C) and do not depend upon the extent of the reaction.
Changing the balancing coefficients in a 1/2 reaction does not change the value of E° or E. HOWEVER,
changing the balancing coefficients will change ∆G° and ∆G (extensive property) as we shall see latter.
Coulomb (C):
Fundamental unit of electric charge.
One electron has a charge of 1.602x10-19 C.
1 C = 6.241x1018 e– = 1.0364x10-5 mole e–
Faraday (F):
Absolute charge of one mole of electrons.
F = 9.6485x104 C/mole
Voltage (V):
The driving force or “electrical pressure”
electrons “feel” to complete a process.
1 V = 1 J/C
Energy (J):
1 J = kg•m2/s2 = 1 V*C
Current (A):
the number of electrons that “flow” per second.
ampere (A or Amp), 1 A = 1 C/s
Work (w):
Voltaic cell (process is spontaneous, system can do work):
wmax = -nFE = ∆G (max work done by system)
Electrolytic cell (process is non-spontaneous, work must be done
on system by surroundings, Eexternal, to drive reaction):
w = nFEext (work done on system)
Watt (W):
electrical power unit. The rate of energy expenditure per second.
1 W = 1 J/s=V•A
1 kWh = 3.6x106 J = (1000 W * 3600 s)
Electrochemistry
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Electrochemistry and Thermodynamics
From Michael Faraday’s1 work on electrochemical cells the difference in free energy (∆G) was found to be
directly proportional to the electrochemical potential of the cell, E. The equations are:
∆G = -nFE (non-standard conditions)
∆G˚ = -nFE˚ (standard conditions)
Using this, an equation can be written that directly relates the standard emf of a redox reaction to its
equilibrium constant.
What does this mean? For redox reactions, we can determine ∆G˚ and K for the reaction if we
know (measure or calculate) the cell potential under standard conditions.
Calculate ∆G° and the equilibrium constant for the disproportionation of copper(I) ion at 25°C. Are copper(I)
ions stable in aqueous solution?
2 Cu+(aq) <—> Cu2+(aq) + Cu(s)
1Faraday,
Michael: The English chemist and physicist Michael Faraday, (b. Sept. 22, 1791, d. Aug.
25, 1867), is known for his pioneering experiments in electricity and magnetism. Many consider
him the greatest experimentalist who ever lived. Several concepts that he derived directly from
experiments, such as lines of magnetic force, have become common ideas in modern physics.
Electrochemistry
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Electrochemistry, Thermodynamics and Non-Standard Conditions
(Nernst Equation)
Changes in concentrations of the electrolytes that make up the half-cells will cause the cell potential, Ecell, to
change. This change can be understood by looking at how changes in concentrations effect the free energy
change of the system. Recall from thermodynamics that the free energy change for the system (∆G) when it
moves from nonstandard-state conditions to equilibrium is given by:
∆G = ∆G˚ + RT ln Q = -RT ln(K/Q)
where Q is the reaction quotient. (Remember the magnitude of Q compared to K can be used to predict the
direction of chemical change to reach equilibrium.) Since ∆G = -nFEcell and ∆G˚= -nFE°cell we can derive an
equation (The Nernst Equation2) that allows us to calculate the cell potential under nonstandard conditions.
Let’s derive this equation!
2
Nernst, Walther: The German physical chemist and inventor Hermann Walther Nernst, b. June 25,
1864, d. Nov. 18, 1941, was awarded the NOBEL PRIZE for chemistry in 1920 for his discovery (1906)
of the third law of thermodynamics, which states that entropy approaches zero as temperature
approaches absolute zero. He introduced the Nernst equation (1889), which relates electric potential
to various properties of the electric cell.
Electrochemistry
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“Think About it”: Electrochemistry and Thermodynamics
Complete the table on the right below by: (1) giving the sign of ∆G°, (2)
indicating if K is less than, greater than or equal to 1 and (3) determining if
the forward reaction is spontaneous, nonspontaneous or at equilibrium.
Complete the table on the left below by: (1) giving the sign of ∆G, (2)
indicating if Q is less than, greater than or equal to K, and (3) determining if
the forward reaction is spontaneous, nonspontaneous or at equilibrium.
∆G˚
(Q = ?)
K
E˚cell
Spontaneous
Forward
Reaction?
∆G
K vs
Q
Ecell
>0
>0
0
0
<0
<0
Spontaneous
Forward
Reaction?
Electrochemistry
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Using the Nernst Equation
Consider the following STANDARD voltaic cell:
Pt|H2 (g,1.0 atm) | H+(aq,1.0 M) || Fe3+ (aq,1.0 M), Fe2+ (aq,1.0 M) | Pt
ANODE
CATHODE
1.
Write the balanced net ionic redox equation for the reaction
2.
Without doing ANY calculations, determine if the reaction is spontaneous under the conditions given.
3.
Determine E°, ∆G° and the equilibrium constant, K.
4.
What is E under the following conditions:
[Fe3+] = 1.50 M, [Fe2+] = 0.0010 M, P(H2) = 0.50 atm and the pH in both half cells is 5.00?
Electrochemistry
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Electrochemistry and Thermodynamics
1. Consider the following redox reaction at 25°C:
1.1. Determine E° = +0.43 V
n=4
4 Ag(s) + O2(g) + 4 H+(aq) <-> 4 Ag+(aq) + 2 H2O(l)
1.2. ∆G° = –166 kJ
1.3. K = 1.4x1029
1.4. If [Ag+] = 1.0 M, the pH = 7.00, and the over pressure of O2 is 0.20 atm, find E = 0.005 V
2. Now consider this redox reaction:
2 Ag(s) + 1/2 O2(g) + 2 H+(aq) <-> 2 Ag+(aq) + H2O(l)
2.1. Determine E° = ?
2.2. ∆G° = ?
2.3. K = ?
2.4. If [Ag+] = 1.0 M, the pH = 7.00, and the over pressure of O2 is 0.20 atm, find E = ?
Electrochemistry
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Using the Nernst Equation
Consider a cell that is constructed based on the following reaction at 25°C:
Sn2+(aq) + Pb(s) –> Sn(s) + Pb2+(aq)
1.
Determine E° for the reaction in the direction it is written.
2.
A voltaic cell is constructed using the reaction shown above. The concentration of Sn2+ in the cathode half-cell
is 1.00 M. The anode compartment contains [SO42−] = 0.50 M and an unknown concentration of Pb2+ ions in
equilibrium with PbSO4(s) at 25°C. Under these conditions, the cell generates an emf of +0.220 V. Determine
Ksp for PbSO4.
Electrochemistry
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Using the Nernst Equation
A voltaic cell utilizes the following reaction:
2 H+(aq) + Zn(s) –> H2(g) + Zn2+(aq)
What is the effect on the cell emf of each of the following changes?
1.
Water is added to the anode half-cell, diluting the solution.
2.
The size of the zinc electrode is increased.
3.
A solution of Zn(NO3)2(aq) is added to the anode compartment, increasing the moles of Zn2+, but not
changing its concentration.
4.
Sodium hydroxide is added to the cathode half-cell.
5.
What happens to the cell voltage as the cell operates?
Electrochemistry
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A Consequence of the Nernst Equation: Concentration Cells
An examination of the Nernst equation shows that E can be non zero even if E° is zero. Why? E will be non
zero if Q ≠ 1. This allows us to construct a concentration cell. In a concentration cell the half-reactions occurring
at the anode and cathode are the reverse of each other but the aqueous ion concentrations are unequal between
the two half cells cells.
Consider the following cell: Ni(s) | Ni2+(aq) || Ni2+(aq) | Ni(s)
E˚ is zero:
E˚ = E˚red + E˚ox = -0.28 V + (+0.28 V) = 0.00 V.
However, if the concentrations of the Ni2+ ions are different in the half-cells, then a cell
potential exists, E, and electricity will flow until the Ni2+ concentrations in the half-cells become equal.
Determine E at 25°C for the
concentration cell depicted on the
left.
“Think about it”: What are the signs
of ∆G°, and ∆G for a concentration
cell?
Electrochemistry
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Concentration Cell Problem
Consider a concentration cell at 25°C using Ag(s) electrodes and AgNO3(aq) as the electrolyte. Let the concentration of
Ag+ ions at electrode A be 0.010 M and the concentration of Ag+ ions at electrode B be 0.50 M.
1.
In what direction will electrons flow to equalize the Ag+ ion concentrations between the two cells? Which
electrode corresponds the the anode half-cell? the cathode? (Sketching a drawing of the cell is helpful here.)
2.
Write the half reaction for each electrode and the overall cell reaction.
A:
B:
Overall cell reaction:
3.
Define and calculate Q for this system.
4.
Calculate Ecell and ∆G for this concentration cell.
5.
“Think about it”: What are the signs of ∆Hsys, ∆Ssys, ∆Ssurr, ∆Suniv and ∆G for this process?
Electrochemistry
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Measuring pH
A pH meter employs a voltaic cell for which the cell potential is very sensitive to pH. A simple (but impractical) pH meter
can be constructed by using two hydrogen electrodes: one standard hydrogen electrode (H2 = 1 atm [H+] = 1.00 M) and
another electrode (H2 = 1 atm [H+] = ?) dipped into the solution of unknown pH > 0. The two half-cells are connected by
a salt bridge or porous glass disk. This is essentially a concentration cell for H+ ions.
1.
If a solution produces an emf of 0.293 V at 25°C with this “pH” probe, what is the pH of the unknown solution?
Electrochemistry
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Electrochemistry, Equilibria and Thermodynamics
The standard 1/2 cell potential for the reduction of the ppt AgSCN(s):
AgSCN(s) + e– —> Ag(s) + SCN–(aq)
E° = +0.0895 V
1. Combine this reduction half-reaction with an oxidation half-reaction to obtain the chemical reaction for the Ksp of
AgSCN(s).
2. Determine E° and the value of Ksp at 25°C for this system.
Note: What we have done is written a non-redox reaction, a Ksp reaction, as a sum of redox half-reactions! This is
not common but can be done for some systems to determine equilibrium constants.
Electrochemistry
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Using the Nernst Equation for 1/2 cell rxns
The Nernst equation can be applied to a half-cell reaction to correct a half-cell reduction (or oxidation) voltage for
nonstandard conditions. Consider the reduction of permanganate ions in acid condition:
MnO4–(aq) + 8 H+(aq) + 5 e- –> Mn2+(aq) + 4 H2O(aq)
1.
E°red = 1.51 V
If the reduction of 1.0 M MnO4– is carried out in a pH = 2.00 buffer with 1.0 M Mn2+ , what is Ered? (Note: under
standard conditions pH = 0.00 = 1 M [H+].)
E = Eo −
⎞
RT ln(Q)
0.0257Vmol ⎛
[Mn 2+ ]
= 1.51V −
ln ⎜
2−
+ 8⎟
nF
5mol
⎝ [MnO4 ][H ] ⎠
1.1. Repeat the calculation for a pH = 7.00 buffer. How sensitive is this reduction voltage to pH change?
Electrochemistry
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