Shorthand Notation for an Electrochemical Cell.

To avoid having to sketch an electrochemical cell each time, a shorthand method of indicating the cell has been formulated.

Starting at the anode and passing through the cell, indicate the cell components by their chemical formulae. A phase change is indicated by a vertical line: |. A salt bridge (or similar device for connecting internally two half reactions) is indicated by two vertical lines: ||.

Example 1:

Calculate Eºcell for: Al | Al^{3+}(1.00 M) || Ag^{+} (1.00 M) | Ag.

From standard reduction potential tables:

anode/oxid | Al ¾ Al^{3+} + 3 e^{-} |
1.66 v |

cathode/reduct | Ag^{+} + e^{-} ¾ Ag |
0.80 v |

balanced equ | 3 Ag^{+} + Al ¾ 3 Ag + Al^{3+} |
2.46 v |

Remember that the cell potential does not depend on the amount of reaction occurring so do not triple the voltage for the silver half reaction, even though you must triple the half reaction to obtain a balanced chemical equation.

Example 2:

Sketch the cell: Ni | Ni^{2+} (1.00 M) || Cu^{2+} (1.00 M) | Cu.

Answer:

Nickel is the anode, copper the cathode in this cell.

**The variation of E _{cell} with Activity (Concentration).**

As mentioned earlier, E_{cell} does not vary with the amount of chemicals involved, but it
does vary with the concentration of the ions (and the pressure of gases) involved in the
cell. It also varies with temperature.

In short, E_{cell} has the characteristics of a typical thermodynamic function. To relate it
to the thermodynamic functions we have met, we must look at the work terms.

DG is the free energy change in a reaction, and it is also the maximum amount of work which a change can produce:

DG = -w_{max}.

Take care with the sign: work produced by a system equates with an energy loss by the system.

The question now is, **What is the work done by an electrical current?**

From physics, the electrical work is equal to the charge, q, times the voltage, E_{cell} (in
our case).

For us the charge, q, is given by the charge on the electron times the number of electrons passing, or better, the charge on a mole of electrons times the number of moles of electrons passing.

The charge on one mole of electrons is called the Faraday, symbol F, and is equal to 96,485 C.

Electrical work then is:

qE = nFE_{cell}.

n is the number of moles of electrons transfered,

F is the charge per mole of electrons, = 96,500 C,

E_{cell} is the cell voltage.

Equating the two work terms:

DG = -nFE_{cell}.

**Thermodynamics and the electrochemical cell.**

Since DG = -nFE_{cell}, substitution into the various thermodynamic expressions
involving DG give the following:

1. From DG_{T} = DGº_{T} + RT ln Q

comes -nFE_{cell} = -nFEº_{cell} + RT ln Q

or, since T usually is taken to be 298 K:

E_{cell} = Eº_{cell} - {R*298 / nF} ln Q

E_{cell} = Eº_{cell} -{0.0257 / n} ln Q

Note that this relationship is often quoted in terms of log_{10}:

E_{cell} = Eº_{cell} - {0.0592 / n} log Q

This relationship is known as the Nernst equation, and it allows the calculation of a the potential for a cell operating under non-standard conditions of activity.

Important to remember in the Nernst equation:

**n refers to the number of electrons transferred during the reaction expressed by
the balanced chemical equation.**

2. From DGº_{T} = -RT ln K

comes -nFEº_{cell} = -R*298 ln K

or, at 298 K:

Eº_{cell} = {0.0257 / n} ln K

or Eº_{cell} = {0.0592 / n} log_{10} K

This equation allows the calculation of K from Eº_{cell}, or vice versa.

**Use of the Nernst Equation.**

Calculate the cell potential for the voltaic cell:

Pb | Pb^{2+} (0.500 M) || Fe^{2+} (0.222 M), Fe^{3+} (0.455 M) | Pt.

**Method**:

Since the cell is a non-standard cell, the Nernst equation is required. This means calculating Eºcell and Q. To find these we need to know the balanced chemical equation and the half reactions.

**Solution**:

From the table of standard reduction potentials:

Fe^{3+} + e^{-} ¾ Fe^{2+} |
Eº = 0.77 v |

Pb ¾ Pb^{2+} + 2 e^{-} |
Eº = 0.13 v |

2 Fe^{3+} + Pb ¾ 2 Fe^{2+} + Pb^{2+} |
Eº_{cell} = 0.90 v |

n = number of electrons transferred = 2

Q = { [Fe^{2+}]^{2} *[Pb^{2+}] } / { [Fe^{3+}]^{2} } (no units, these are activities)

Q = (0.222)^{2} * (0.500) / (0.455)^{2} = 0.119

E_{cell} = Eº_{cell} - {0.0257 / n} ln Q

E_{cell} = 0.90 v - {0.0257 / 2} ln 0.119 = 0.90 v + 0.027 v

E_{cell} = 0.93 v

**Calculation of an Equilibrium Constant from Reduction Potentials.**

Calculate the equilibrium constant for the reaction:

Pb^{2+} + Ni ¾ Pb + Ni^{2+}

Method: K is related to Eºcell by the relationship:

Eº_{cell} = {0.0257 / n } ln K

Solution:

Pb^{2+} + 2 e^{-} ¾ Pb |
-0.13 v |

Ni ¾ Ni^{2+} + 2 e^{-} |
+0.25 v |

Pb^{2+} + Ni ¾ Pb + Ni^{2+} |
0.12 v |

Substituting:

0.12 = {0.0257 / 2} ln K

ln K = 9.34

K = 1.14 x 10^{ 4}

(no units, this is the activities K.)

Calculation of Equilibrium Position.

For the following cell, starting concentrations shown, deduce the concentrations of all ions when the reaction comes to equilibrium.

Pt | Fe^{2+} (0.500 M), Fe^{3+} (0.100 M) || Ag^{+} (0.400 M) | Ag

Method: Calculate K from Eº_{cell}, calculate equilibrium position using ICE for the
analysis.

Solution:

oxidation/anode | Fe^{2+} ¾ Fe^{3+} + e^{-} |
-0.77 v |

reduction/cathode | Ag^{+} + e^{-} ¾ Ag |
0.80 |

balanced equation | Fe^{2+} + Ag^{+} ¾ Fe^{3+} + Ag^{} |
0.03 v |

ln K = n * Eºcell / 0.0257 = 1 * 0.03 / 0.0257 = 1.167

K = 3.21

Having calculated K, use ICE to analyse the equilibrium part of the problem:

Fe^{2+} |
+ Ag^{+} |
¾ | Fe^{3+} |
+ Ag | |

I (M) | 0.500 | 0.400 | 0.100 | some | |

C | (-1)x | (-1)x | (+1)x | (+1)x | |

E | 0.500 - x | 0.400 - x | 0.100 + x | some |

K = (0.100 + x) / {(0.500 -x)(0.400 - x)} = 3.21

Solve for x: x = 0.1607

So at equilibrium:

[Fe^{2+}] = 0.339 M;

[Fe^{3+}] = 0.261 M;

[Ag^{+}] = 0.239 M.

Using Reduction Potentials to Calculate Equilibrium Constants for non-Redox Reactions.

Although this may seem to be impossible, in fact it is readily done for, for example, solubility equilibria.

From reduction potential tables calculate K_{sp}(AgI).

Method: the "secret" here is to realize that the reduction half reaction and the oxidation half reaction must be for the same material, so that when they are added to give the balanced chemical equation, there is no redox!

oxidation | Ag(s) ¾ Ag^{+}(aq) + e^{-} |
-0.80 |

reduction | AgI(s) + e^{-} ¾ Ag(s) + I^{-}(aq) |
-0.15 |

balanced | AgI(s) ¾ Ag+(aq) + I^{-}(aq) |
-0.95 |

Although the balanced equation is not a redox, an Eº_{cell} has been calculated for it! And
from this, K can be calculated.

Eº_{cell} = {0.0257 / n} ln K

-0.95 = {0.0257 / 1} ln K

ln K = -36.96

K = 8.84 x 10^{-17}.

The Range of K values calculable from Eº_{cell}.

As noted in the above problem, a small negative voltage (less than 1 v, which is easily measurable) gave a very small K value. Consequently it is possible to measure extremely small (or large) equilibrium constant values using electrochemical methods.

The largest Eºcell found is about 6 v. What equilibrium constant does this lead to?

Eº_{cell} = {0.0257 / n} ln K

6 = {0.0257 / 1} ln K

ln K = 233.5

K = 2.45 x 10^{101}.

The range of equilibrium constants available from electrochemical measurements is in the approximate range

K = 10^{-100} to 10^{+100}