Download Lecture Oct 29, 2007 – Chapter 14 – Electrochemistry

Lecture Oct 29, 2007 – Chapter 14 – Electrochemistry
Electrochemical Cells
-
V
+
Ag(s)
Cd(s)
Plated with AgCl(s)
0.0167 M
CdCl2(aq)
Anode
Cathode
This cell has the POTENTIAL to do work due to the voltage of the cell.
How does it do work???
e-
V
+
eAg(s)
Cd(s)
Plated with AgCl(s)
0.0167 M
CdCl2(aq)
Anode
Cathode
It does work by electrons flow through an outer wire! To what extent do the
electrons want to flow from the anode to the cathode in order to reduce Ag+
In this electrochemical cell, Cd is being oxidized and the Ag+ from AgCl(s) is being
reduced at the Ag elecrtode.
AgCl(s) + eCd(s)
Ag(s) + Cl- (aq)
Cd2+(aq) + 2e-
2AgCl(s) + Cd(s)
2Ag(s) + 2Cl-(aq) + Cd2+(aq)
REMINDER!
We can describe the work on the surroundings as delta G or Gibb’s free energy.
Delta G = - work = -Eq ;
Where E is equal to the cell potential and q is equal to the charge on an electron
E is in units of V or J/C
q is in units of C or coulombs
therefore
G is in units of Joules or J
NOTE THAT
q = nF
THEREFORE
Delta G = -nFE
n = moles of electrons that are invovled in the oxidation-reduction reaction!!!!
F = Faraday constant or 9.649 x 10^4 C/mole e- (comes from 1.602 x 10^-19 C * 6.022 x 10^23 mol-1)
Delta G for a SPONTANEOUS reaction is NEGATIVE
REMEMBER THAT
DeltaG = DeltaH - TDeltaS; where deltaH is enthalpy and deltaS is entropy
A negative deltaG results from a negative deltaH (heat given off) and a positive deltaS
(disorder)
IF deltaG for the above cell reaction is -150 KJ/mol Cd (driving force for reaction) then
what is the cell potential in V?
DeltaG = -nFE
SO,
E = deltaG/(nF)
E = +0.777 V
A negative deltaG results in a positive cell potential thereby describing the cell’s
potential to DO WORK!
The greater the voltage, the more current will flow
The greater the resistance, the less current will flow
REMEMBER that I = V/R
Why will the following cell not WORK or why is the potential to do work for the
following cell low?
+
V
Ag(s)
Cd(s)
Cd(NO3)2 (aq)
AgNO3 (aq)
Anode
Cathode
Because Ag+ in solution are what needs to be reduced and they are NOT FORCED
to be reduced at the Ag electrode. There is nothing preventing them from being
reduced at the Cd electrode.
NEED for electrons to flow through outer wire that is connected to the anode
electrode and cathode electrode through which the voltage is measured by a
potentiometer.
HOW DO WE FORCE THE Ag+ to be reduced at the Ag electrode???
With a SALT BRIDGE in a spontaneous electrochemical cell
e-
Cl-
Cd(s)
Cd2+
CdCl2 (aq)
Anode
V
e-
+
K+
Ag(s)
Ag(s)
AgNO3 (aq)
NO3Cathode
Salt Bridges are made out of KCl and contain sintered glass frits (semipermeable
barrier) at the openings which allow for ion migration to offset charge buildup due
to flow of electrons into cathode. The salt bride forces Ag ions to be reduced at the
Ag(s) electrode.
NOTE that KCl or any other electrolyte not involved in the reaction can be used!