Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 110

General Chemistry

Summer 2006

Lecture Notes::Lec 9_9 June

© R. Paselk 2006
 
     
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Galvanic Cells, cont.

Anodes vs. Cathodes.

Electrodes and Electrode Potentials

Reference Electrodes: there is no absolute standard for potential--it must always be defined in reference to something (think about gravitational potential or field), potential can be positive or negative. Thus arbitrarily chose a particular cell as the standard, define its potential, and compare everything else to it.

Standard Reduction Potentials

Let's look again at a Galvanic Cell (note that Zumdahl's illustrations assumes a porous disk connection)

The Galvanic cell consists of two half cells. In each cell the reaction, by convention, is written as a half-reaction, which is in fact the chemistry taking place in the half cells.

Note that in tables of Reduction Potentials [overhead] the reactions are written as reduction half reactions, with the potentials those which would occur if the half-cell were connected into a galvanic cell with a SHE.

For the Zn, Cu, system we then have

Zn half cell:

Zn 2+ + 2e - Zn0 E° = -0.76 V

But of course in our cell, this goes backwards:

Zn0 Zn 2+ + 2e - E° = +0.76 V


Cu Half Cell:

Cu2+ + 2e - Cu0 E° = +0.34 V


CELL :

Cu2+ + Zn0 Cu0 + Zn 2+ E° = +1.10 V


But recall we also stated that these half-cell potentials are always related to the SHE, which is defined as 0.00 V at all temperatures.

The voltages of half cells determined relative to SHE are the Standard Reduction Potentials for these half cells (1M [more properly, at an activity of 1], 1 ATM)

So how do we determine the V of a Galvanic Cell from half-cell voltages?

1. Electrons will flow from less positive to more positive cell.

Consider a Galvanic Cell of Ag+/Ag and Cu2+/Cu


What will the reaction be?


Ag+ + e - Ag0 E° = +0.80 V

Cu2+ + 2e - Cu0 E° = +0.34 V


Since e- flow toward the more positive half cell, then Ag+/Ag cell is cathode (reduction at cathode) and Cu2+/ Cu will be reversed.

Now we need to balance electrons so

2 (Ag+ + e - Ag0) E° = +0.80 V


Cu0 Cu2+ + 2e - E° = -0.34 V



2 Ag+ + Cu0 2 Ag0 + Cu2+ E° = + 0.46 V


Cell Diagrams

Instead of drawing cells we often draw a cell diagram using what your author refers to as "Line Notation." It is conventional to start with the anode on left. The cell diagram for the Zn/ Cu Galvanic Cell will then be represented as:

Zn | ZnSO4(x M) | CuSO4(y M) | Cu

or Zn | Zn2+(x M)| Cu2+(y M) | Cu

Both of these describe cells with a liquid junction, such as a porous disk. What do Lines represent? Changes in phase or boundaries. With a salt bridge we see:

Zn | ZnSO4(x M) || CuSO4(y M) | Cu

Note the two lines in the center representing the salt bridge. We need two lines because there is a boundary between each end of the salt bridge and its respective half cell.

So the complete description of our cell includes:

  1. The cell potential (= 1.10 V).
  2. The electron flow direction.
  3. Designation of Anode and Cathode. (Note that these terms can refer to the metal probe or the half cell. Note also that the term electrode often refers to the entire half cell.)
  4. The nature of each electrode, and the solution in each half cell.

Cell Potential and deltaG

If we think physics, then the work done by a system is:

w = -qE

Where E = potential difference and q = charge.

So for the system or reaction:

-w = qE

and

q = nF

where F = 96,485 coulombs/mole = the Faraday constant. Note also that E will change (decrease) if currrent is flowing.

If we consider a system doing the maximum possible work (no currrent flow - so takes forever), then work equals free energy, or:

deltaGmax = deltawmax = -qEmax =-nFEmax

For standard conditions @ equilibrium:

deltaG° = -nFE°

deltaG = deltaG° + RT ln Q

Substituting -nFE for deltaG:

-nFE = -nFE° + RT ln Q

Dividing by -nF:

E = E° - (RT/nF) ln Q

This is called the Nernst Equation, which relates the voltage produced by a cell to the concentrations of reactants and products in a system. Note that this is the electrochemical version of the equation for free energy in a non-equilibrium system.

or, substituting values for R, T, and F at 25°C:

E = E° - (0.0257/n) ln Q

We also noted the relationship between free energy and voltage under standard conditions:

deltaG° = -nFE°

So let's use this relationship in a typical problem.

Example: Calculate the voltage for the cell:

Sn(s)| Sn2+(aq) || Ag+(aq) | Ag(s)

@ 25°C assuming Sn 2+(aq) = 0.15 M and Ag +(aq) = 0.30 M.

Standard potentials for the half reactions are:

Sn2+(aq) + 2e - Sn(s) E°= -0.14 V

Ag+(aq) + 2e - Ag(s) E°= + 0.80 V

Electrons will flow to the more positive half cell, where reduction takes place. Therefore the Ag +/Ag half cell is the cathode, and the Sn2+/Sn half cell must be reversed to become the anode:

Sn(s) Sn2+(aq) + 2e - E°= +0.14 V

2(Ag+(aq) + 2e - Ag(s)) E°= + 0.80 V


2 Ag+(aq) + Sn(s) Ag(s) + Sn2+(aq) E°= + 0.94 V

For this reaction the mass action expression, Q, is Q = [Sn2+]/[Ag+]2.

Substituting into the Nernst equation:

E = E° - (RT/nF) ln Q

E = 0.94 V - (0.0257/2) ln [0.15]/[0.30]2

E = 0.94 - (-0.006564) = 0.947 V = 0.95 V

Concentration Cells

Concentration cells are cells where the components in the two half cells are identical, but the concentrations differ. So what will the voltage of such a cell be? Consider a cell made up with 0.10 M Ag+ on one side and 1.0 M Ag+ on the other.

Ag+(aq) + 2e - Ag(s) E°= + 0.80 V

Ag(s) Ag+(aq) + 2e - E°= - 0.80 V

Notice that the E° values cancel to give a standard potential of 0.00 V. The question is, which is the anode and which is the cathode?

Let's look at the system qualitatively and see if we can reason which half cell is which.

So the system should be written as:

0.10 M = anode: Ag(s) Ag+(aq) + 2e - E°= - 0.80 V

1.0 M = cathode: Ag+(aq) + 2e - Ag(s) E°= + 0.80 V

or, as a cell diagram, with the anode on the left:

Ag(s) | Ag+(aq) = [0.10 M] || Ag+(aq) = [1.0 M] | Ag(s)

We can now use the Nernst equation to calculate the voltage:

E = E° - (RT/nF) ln Q

E = 0.00 V - (0.0257/1) ln [0.10]/[1.0]

E = 0.059 V

Note that V is very small (59 mV for a change in concentration of 10). We can use such changes of voltage with concentration as a way of determining concentration called potentiometry.

Potentiometry

Potentiometry refers to the use of potential differences to determine concentration differences. The most common and familiar use of potentiometry is in pH meters.

Let's look a bit at how we might determine pH (that is [H+]) by potentiometry.

Reference Electrodes

Calomel (SCE). One of the most common reference electrodes is the so called calomel electrode (named for the calomel filling = Hg2Cl2)

Diagram (note the diagram is for the half cell only):

Hg | Hg2Cl2 (sat'd), KCl (x M) ||

for the half-reaction:

Hg2Cl2(s) + 2e - 2 Hg(l) + 2 Cl-(aq)


Note that there is only one variable contributing to the Nernst equation, the concentration (activity) of chloride ion, since calomel and mercury are in the pure states and thus have activities of 1! Thus the potential is:

  E = E° ­ (RT/nF) ln Q

@ 25°C, E = E° - (0.0257/1) ln [Cl-] = 0.224 V

for the saturated version of the Standard Calomel Electrode (SCE). The SCE is very common because its easy to prepare and very stable, and it has a well defined potential.

The calomel electrode has lost some popularity due to its mercury content. The most common alternate electrode is the Ag/AgCl reference electrode, which we will not discuss.

So now we have a reference electrode, what about the electrode for measuring hydrogen ion? Again the SHE type electrode is difficult to work with and is very rarely used.

The most common electrode for measuring pH is the glass electrode. (overhead - see figure in text.)


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