Electrochemistry
The standard cell potential for the Daniell Cell is `E_("cell")^o =1.10V`
Calculate the free energy change, `DeltaG^o`, the equilibrium constant, `K` and comment on the thermodynamic viability.
`DeltaG^o=-nFE_("cell")^o`
`E_("cell")^o=E_("red")^o + E_("ox")^o`
`E^o` are such that they refer to reduction reactions ie.,
`Cu^(2+)+2e^(-)rarrCu; E^o=+0.34V`
`Zn^(2+)+2e^(-)rarrZn; E^o=-0.76V`
The reaction with largest `E^o` determines the forward reaction, ie.,
|
`ZnrarrZn^(2+)+` |
`2e^(-)` |
| `2e^(-)+Cu^(2+)rarrCu` |
Overall we have,
`Zn+Cu^(2+)rarrZn^(2+)+Cu`
For which the cell potential can now be expressed,
`E_("cell")^o=E_("red")^o +E_("ox")^o=0.76V+0.34V=1.10V`
Using the Nernst equation, we can now calculate the Gibbs free energy,
| `DeltaG^o` | `=-nFE^o` |
| `=-2xx96,485Cmol^(-1)xx1.10V` | |
| `=212,267Jmol^(-1)` |
From the Gibbs free energy change, we can calculate the Equilibrium constant,
`DeltaG^o=-RTlnK`
`(DeltaG^o)/(-RT)=lnK=85.68`
`K=exp(lnK)=1.62xx10^37`
Since `K">>"1`, the reaction goes far towards completion, and since `DeltaG^o` is negative, the reaction is spontaneous.
Dependence of Reduction Potentials on Cell Conditions
Positive `E^o` means the reaction is favourable, but just by looking at `E_("cell")^o` it is not always possible to determine which reaction will occur. The lowest valued `DeltaG^o` will go. Previous analysis focussed on standard reduction potentials, ie., at `T=298K`, `P=1"bar"` and `"conc"=1moldm^(-3)`, but in the lab, this is rarely the case.
The Nernst Equations and Concentration
For example, take the reduction of Zn2+.
`Zn^(2+)+2e^(-)earr Zn; E^o=0.76V`
When the concentration, [Zn2+] is low, the equilibrium will shift towards Zn2+. So we can expect `DeltaG` to be less negative, and E to be more negative.
`E=E^o-(RT)/(nF) lnQ`
For this couple, we have,
`Q=1/([Zn^(2+)])`, and `n=2`
So we can calculate the actual couple voltage,
`E=-0.76-(RT)/(nF) lnQ`
Consider the example here [Zn2+]=0.1moldm-3,
|
`E` |
`=-0.76V-(8.314xx298)/(2xx96485) xxln (1/0.1)` |
| `=-0.79V` |
E is less positive under these conditions and thus demonstrates the reaction to be less favourable.
Dependance of Voltage on pH
Consider the Manganate reduction
`MnO_4^(-)+8H^(+)+5e^(-)earrMn^(2+)+4H_2O; E^o=1.55V`
For this reaction, when we express the equilibrium constant, we see that this involves the proton concentration,
`K=([Mn^(2+)])/([MnO_4^(-)][H^(+)]^8`
Let [Mn2+] and [MnO4-] be of unit concentration, and we can see the strong pH dependance,
`K=1/([H^(+)]^8)`
Where the proton concentration can be expressed as the anti-cologarithmn of pH,
`[H^(+)]=10^(-pH)`
Hence,
`K=1/(10^(-8pH))`
We can express the Gibbs free energy change for the redox reaction,
`DeltaG=-RTlnK=-RTxx8pHxxln10`
In turn, we can express the redox potential in terms of the Gibbs free energy, so,
`E=E^o-(RT)/(nF)xx8pHxxln10`
| pH | E/V | |
| 1 | 1.46 | |
| 7 | 0.89 | |
| 14 | 0.22 |
With increasing pH, E becomes less favourable and hence explain why MnO4- will oxidise Cl- in acid, but not under neutral conditions.