Core practical 10 - construct electrochemical cells and measure electrode potentials

Calculate E^ө [Fe²⁺_(aq) | Fe_(s)].

E [Fe(s) | Fe2+(aq)] and [Cu2+(aq) | Cu(s)] = 0.78V (or your own results).

E cell = E _{right-hand half-cell} − E _{left-hand half-cell}

0.75 = 0.34 − E [Fe²⁺_(aq) | Fe_(s)]

The value for the iron half-cell is −0.44V.

The electrode potential values for the cells you set up may be slightly different to theoretical values. Give a reason for this.

The experiment was not carried out under standard conditions.

Give a reason why silver nitrate is not used as a 1.0 mol dm⁻³ solution

Silver nitrate is highly oxidising; an alternative answer is that silver nitrate is very expensive.

[Mg²⁺_(aq) | Mg_(s)] can also be used as a half-cell. Describe a problem that might be observed with this system.

Magnesium reacts slowly with the water in the solution, raising the concentration of magnesium ions. The equilibrium will move to oppose this change and form more magnesium atoms.