Download Applications of standard potentials

Midterm Exam 1: Feb. 2, 1:002:10 PM at Toldo building,
Room 100
Example: Calculate emf of the cell : Mn(s)|Mn+2||Fe+3 , Fe+2|Pt(s)
Solution: The two reduction half reactions
Right (cathode):
Fe+3(aq) + e- → Fe+2(aq)
Left (anode):
Mn+2(aq) + 2e- → Mn(s)
It shows that the above two half reactions have different numbers of electrons
transferred
The cell reaction is obtained through 2*R – L,
2Fe+3(aq) + Mn(s) → 2Fe+2(aq) + Mn+2(aq)
should the cell potential be calculated as 2Eright - Eleft ?
Answer: ???
it leads to
ΔrG = 2ΔrG(R) - ΔrG(L)
2FE = 2*1*F* E(R) – 2*F* E(L)
Ecell = E(R) - E(L)
For the standard cell potential:
Eөcell = 0.769 - (- 1.182) = 1.951 V
Standard Cell emf
•
Eθcell = Eθ (right) – Eθ(Left)
•
Calculating equilibrium constant from the standard emf :
Evaluate the solubility constant of silver chloride, AgCl, from cell potential
data at 298.15K.
Solution:
AgCl(s) → Ag+(aq) + Cl-(aq)
Establish the electrode combination:
(Assume this one)
Right: AgCl + e- → Ag(s) + Cl-(aq) Eθ = 0.22V
(Obtain this one through C – R) Left: Ag+(aq) + e- → Ag(s)
Eθ = 0.80V
The standard cell emf is : Eθ (right) – Eθ(Left) = - 0.58V
vFE
 0.58
ln K 

RT
0.0257
K = 1.6x10-10
•
The above example demonstrates the usefulness of using two half reactions
to represent a non redox process. What would be the two half reactions for
the autoprotolysis of H2O?
The measurement of standard
potentials
• The potential of standard hydrogen electrode:
Pt(s)|H2(g)|H+(aq)
is defined as 0 at all temperatures.
• The standard potential of other electrodes can be obtained by
constructing an electrochemical cell, in which hydrogen electrode is
employed as the left-hand electrode (i.e. anode)
• Example: the standard potential of the AgCl/Ag couple is the
standard emf of the following cell:
Pt(s)|H2(g)|H+(aq), Cl-(aq)|AgCl(s)|Ag(s)
or
Pt(s)|H2(g)|H+(aq) || Cl-(aq)|AgCl(s)|Ag(s)
with the cell reaction is: ½ H2(g) + AgCl(s) → H+(aq) + Cl-(aq) + Ag(s)
The measurement of standard
potentials
• The potential of standard hydrogen electrode:
Pt(s)|H2(g)|H+(aq)
is defined as 0 at all temperatures.
• The standard potential of other electrodes can be obtained by
constructing an electrochemical cell, in which hydrogen electrode is
employed as the left-hand electrode (i.e. anode)
• Example: the standard potential of the AgCl/Ag couple is the
standard emf of the following cell:
Pt(s)|H2(g)|H+(aq), Cl-(aq)|AgCl(s)|Ag(s)
or
Pt(s)|H2(g)|H+(aq) || Cl-(aq)|AgCl(s)|Ag(s)
with the cell reaction is: ½ H2(g) + AgCl(s) → H+(aq) + Cl-(aq) + Ag(s)
• The Nernst equation of the above cell reaction is
aH  aCl 
RT

EE 
ln{
}
 1/ 2
vF
( f H2 / p )
Using the molality and the activity coefficient to represent the activity:
E = E θ – (RT/vF)ln(b2) - (RT/vF)ln(γ±2)
Reorganized the above equation:
E + (2RT/vF)ln(b) = E θ - (2RT/vF)ln(γ±)
Since ln(γ±) is proportional to b1/2, one gets
E + (2RT/vF)ln(b) = E θ - C* b1/2, C is a constant
Therefore the plot of E + (2RT/vF)ln(b) against b1/2 will yield a straight
line with the interception that corresponds to E θ
Example plot from the text book
(the interception corresponds to E θ)
Example: Devise a cell in which the cell reaction is
Mg(s) + Cl2(g) → MgCl2(aq)
Give the half-reactions for the electrodes and from the
standard cell emf of 3.00V deduce the standard potential of the
Mg2+/Mg couple.
Solution:
the above reaction indicates that Cl2 gas is reduced
and Mg is oxidized.
Therefore,
R:
Cl2(g) + 2e- → 2Cl-(aq) (Eө = + 1.36 from Table 10.7)
L:
Mg2+(aq) + 2e- → Mg(s) (Eө = ? )
The cell which corresponds to the above two half-reactions is :
Mg(s)|MgCl2(aq)|Cl2(g)|Pt
Eөcell = Eө(R) - Eө(L) = 1.36 – Eө(Mg2+/Mg)
Eө(Mg2+/Mg) = 1.36V – 3.00V = - 1.64V
Example: Consider a hydrogen electrode in aqueous HCl solution at 25oC
operating at 105kPa. Calculate the change in the electrode potential when
the molality of the acid is changed from 5.0 mmol kg-1 to 50 mmol kg-1.
Activity coefficient can be found from Atkin’s textbook (Table 10.5 ).
Solution: first, write down the half reaction equation:
H+(aq) + e- → ½ H2(g)
Based on Nernst equation
E  E 
So
RT
ln Q
vF
Q
( PH 2 / P )1/ 2

 1/ 2
RT  ( PH 2 / P ) 
E
ln 

vF 


E2 – E1 = -
RT
ln( 1
Fv
2
) = - 25.7(mV)x ln(
= 56.3 mV
0.929 * 0.005
)
0.830 * 0.05
• Choose the correct Nernst equation for the cell
Zn(s) | Zn2+ || Cu2+ | Cu(s).
A: Δ E = Δ E° - 0.032 log([Zn2+ ]/[Cu2+])
B: Δ E = Δ E° + 0.024 log([Cu2+] / [Zn2+])
C: Δ E = Δ E° - 0.021 log(Zn / Cu)
D: Δ E = Δ E° - 0.018 log(Cu / Zn)
• Answer ...
Hint...
• The cell as written has
Reduction on the Right: Cu2+ + 2e = Cu
oxidation on the left:
Zn = Zn2+ + 2e
• Net reaction of cell is Zn(s) + Cu2+ = Cu(s) + Zn2+
Applications of standard
potentials
The electrochemical series
• For two redox couples Ox1/Red1 and Ox2/Red2,
Red1, Ox1 || Red2, Ox2
Eθ = Eθ2 – Eθ1
The cell reaction: Red1 + Ox2 → Ox1 + Red2
If Eθ > 0 then ΔrGθ < 0 (Nernst equation), the reaction will take
place spontaneously (??). In other words, if Eθ2 > Eθ1, the Ox2 has
the thermodynamic tendency to oxidize Red1.
Which metal is more
Suitable for anode?
Cathode?
The determination of activity
coefficients
• Once the standard potential of an electrode (Eθ)
is known, one can use the following equation
E + (2RT/vF)ln(b) = E θ - (2RT/vF)ln(γ±)
to determine the mean activity coefficient of the
ions at the concentration of interest via
measuring the cell emf (E).
• For example: H2(g) + Cl2(g) → 2HCl(aq)
The determination of equilibrium
constants
• Self-test 7.11 Calculate the solubility constant (the equilibrium
constant for reaction Hg2Cl2(s) ↔ Hg22+(aq) + 2Cl-(aq)) and the
solubility of mercury(I) chloride at 298.15K. The mercury(I) ion is the
diatomic species Hg22+.
• Answer: This chemical process does not involve electron transfer,
i.e. is not a redox reaction.
Choosing cathode reaction as: Hg2Cl2(s) + 2e → 2Hg(l) + 2Cl-(aq)
from reference table 7.2, Eθ = 0.27 V
the anode reaction can be obtained through R – Cell
Hg22+(aq) + 2e → 2Hg(l)
from reference table 7.2, Eθ = 0.79 V
Therefore the standard cell potential = 0.27 – 0.79 = -0.52 V