Electrochemistry II, Cell from Reaction and etc.

Cells with no salt bridge and no liquid junction

The following electrochemical cell has no salt bridge and no junction potential,

Pt(s)|H2(g)|HBr(aq)|AgBr(s)|Ag(s).
(In this cell the AgBr is a coating on the silver electrode. AgBr is insoluble in water, so if you dip silver metal in HBr it forms a thin coating on the metal. The left-hand side is just the standard hydrogen electrode.)

We analyze the cell as follows: (Recall that we write the right hand side as a reduction and, by default, the left hand side as an oxidation.)

R AgBr + e− → Ag + Br −                ERo = 0.0711 v
L ½ H2 → H+ + e−                           ELo = 0
------------------------------------------------------------------
AgBr + ½ H2 → Ag + H+ + Br −
By convention the E o for the cell is given by
Eo = ERo− ELo,
or,
E o = 0.0711 − (0) = 0.0711 v.
Caution must be used when looking in the tables of reduction potentials. The table will show both
Ag+ + e−→ Ag                             0.080 v

and

AgBr + e− → Ag + Br−                0.0711 v.

The second one is the correct half-cell reaction for our problem. The first one cannot be used here because you can't use anything in your reaction that's not in the cell. Thus we must use the half-reaction that has the AgBr and not the one that has the Ag+ because there is no Ag+ showing in the cell diagram.
 
 

One more example to illustrate a point

Consider the cell,

Hg(l)|Hg2Cl2(s)|HCl(aq)|AgCl(s)|Ag(s).
This cell can be analyzed as follows:
R                 AgCl + e− → Ag + Cl−             ERo = 0.22 v
L             2 Hg + 2 Cl− → Hg2Cl2 + 2 e−      ELo = 0.27 v (reduction)
In order for the electrons on both sides to cancel each other out the Ag half-reaction must be multiplied by two before the two half reactions are added together. This gives,
R           2 AgCl + 2 e− → 2 Ag + 2 Cl−            ERo = 0.22 v
L                 2 Hg + 2 Cl− → Hg2Cl2 + 2 e−      ELo = 0.27 v (reduction)
--------------------------------------------------------
             2 AgCl + 2 Hg → Hg2Cl2 + 2 Ag
The Eo for the cell, however, is still given by,
Eo = ERo − ELo = 0.22 − 0.27 = − 0.05 v.
Even though we multiplied the silver half-reaction by two we did not multiply the corresponding half-cell potential by two. That is because each of the three of the Eo's in this problem is essentially a Gibbs free energy per mole of electrons. If we multiplied the Eo for the silver electrode by two mols the units would not match.
 
 

Cell From a Reaction

So far we have been analyzing electrochemical cells by writing the reaction corresponding to a given cell. Now we need to be able to go the other way. We must be able to devise an electrochemical cell which will carry out a given reaction.

If we are given an oxidation/reduction reaction we can devise a cell which will carry out that reaction. For example, consider the reaction,

16 H+ + 2 MnO4−+ 5 Sn2+ → 5 Sn4+ + 2 Mn2+ + 8 H2O.
Mn is being Reduced so all the stuff that contains Mn must be on the Right side of the cell. Likewise Sn2+ is being oxidized so the Sn2+ and Sn4+ ions must be on the left side of the cell.
Pt(s)|Sn2+(aq),Sn4+(aq)|MnO4−(aq),Mn2+(aq)|Pt(s).
This cell has several problems: First, we have two different solutions in contact with each other. This will produce a junction potential. To fix this problem insert a salt bridge, "||," between the two solutions. Second, the reaction is being carried out in acid solution. If you remember your freshman chemistry you will recall that the permanganate ion reduces to Mn2+ in acid solution, but it reduces to MnO2 in basic solution. Likewise, you want the Sn side to be in acid solution so that you won't be in danger of precipitating out one or both of the tin oxides. So, to be safe we should probably write the cell as
Pt(s)|Sn2+(aq),Sn4+(aq),H+(aq)||MnO4−(aq),Mn2+(aq),H+(aq)|Pt(s).


Concentration Cells

It is not always necessary to have a net chemical reaction in order to generate a cell voltage. A concentration cell is an example of a cell which will produce a voltage (albeit a small voltage) without a net chemical reaction. Consider the cell,

Zn(s)|ZnSO4(aq,mL)||ZnSO4(aq,mR)|Zn(s).
(Note that the "L" and "R" refer to the left- and right-hand sides of the cell respectively.) We analyze this cell as follows:
R ( = Reduction)           Zn2+(aq,mR) + 2 e− → Zn(s)
L ( = oxidation)             Zn(s) → Zn2+(aq,mL) + 2 e− .
The overall reaction is just the sum of these two half-reactions:
Zn2+(aq,mR) → Zn2+(aq,mL).
(Note that the zinc ion concentration from the right hand side of the cell appears to the left of the arrow in the balanced overall reaction and vice versa.)

The Eo for this reaction is given, as usual, by,

but in this case the oxidation half reaction is just the reverse of the reduction half reaction so that EoR and EoL are the same and the overall Eo is zero.

Nevertheless, we can generate a voltage if the concentrations on each side are the different. Using the Nernst equation we get

This makes sense, because if the concentrations on the two sides are the same then the voltage is zero. However, if the concentration on the right side of the cell (not the right side of the reaction) is larger then the argument of the logarithm is less than one so that its logarithm is negative and E is positive. We know from other considerations that there is a natural drive toward dilution so that this process is spontaneous.

In principle any isothermal change in Gibbs free energy can be set up to generate a voltage. In this case the Gibbs free energy change in going from a more concentrated solution to a more dilute solution is negative. This negative Gibbs free energy change can be used to give a voltage and produce electrical energy.
 
 

How to measure Eo and γ±

We will show how to measure Eo and γ± by means of an example. Consider the cell

Pt(s)|H2(g, p)|HCL (aq, m)|AgCl(s)|Ag(s)
Analyze it as usual, but now we assume the only emf that we know is that of the standard hydrogen electrode. We want to measure the emf of the AgCl(s)|Ag(s) half-cell. Call it Eo.
R AgCl + e− → Ag + Cl −                ERo = Eo
L ½ H2 → H+ + e−                         ELo = 0
------------------------------------------------------------------
AgCl + ½ H2 → Ag + HCl(aq, m)
In the above reactions the Eo is one of the quantities we want to measure. The left-hand electrode is just the standard hydrogen electrode. We use the Nernst equation approximating H2 as an ideal gas (a pretty good approximation at low pressures).
(1a, b)


Now hold the pressure of H2 constant at 1 atm. Rewrite Equation 1b as,

(2a, b)
Rearrange Equation 2b as follows:
(3)
There are two unknowns in this equation Eo and γ± , but we prepare our solutions so that we know the molality, m, and we measure E at each value of m. Also, we know that γ± → 1 as m → 0.

We find Eo by plotting the quantity  vs m and extrapolate to m = 0 . This curve intercepts the m = 0 axis at Eo.

This might be hard to do in practice because it is tricky to work with very small concentrations. Also, note that ln m gets very large and negative as m gets small. so presumably E is getting large and we are subtracting two large numbers to get a small one.

An improvement can be made by using the Debye-Hückel limiting law.

               (4)


where I = m for HCl(aq) so that

              (5)
Then
               (6a, b)


Remember that E and m are measured.
Plot  vs m and extrapolate to m = 0. The intercept is still Eo, but the extrapolation should be easier to carry out.

Once we know Eo we can obtain γ± at any m by measuring E at that m. From,

                (7)
we solve for ln γ± as,
                (8)
We can use these measurements to make tables of γ± or we can fit the data to a curve.

WRS

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Copyright 2004, W. R. Salzman
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salzman@arizona.edu
Last updated 25 Oct 04
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