Question

Calculate the EMF of the cell:

Al(s) | Al3+ (0.02 M) || Au+ (0.01 M) | Au(s)

Where Al3+ + 3e-  Al E o= -1.66 V

Where Au+ + e-  Au E o = +0.99 V

Answer 1

Al(s) | Al³⁺ (0.02 M) || Au⁺ (0.01 M) | Au(s)

The cell reaction is : Al(s) + Au⁺(aq) Al³⁺(aq) + Au(s)

Remember that always oxidation takes place at anode & reduction takes place at cathode.

Anode reaction : Al(s) Al³⁺(aq)

Cathode reaction : Au⁺(aq) Au(s)

Standard potential of the cell , E^o = E^o_cathode - E^o_anode

                                                 = E^o_Au+/Au - E^o_Al3+/Al

                                                = +0.99 V - (-1.66V )

                                               = 2.65 V

Therefore the cell potential is +2.65 V

According to Nernst Equation ,

E = E^o - (0.059 / n) log ([Products] / [reactants] )

   = E^o - (0.059 / n) log ([Al³⁺] / [Au⁺] )

Where

E = electrode potential of the cell = ?

E^o = standard electrode potential = +2.65 V

n = number of electrons involved in the reaction =3

[Al³⁺] = 0.02M

[Au⁺] = 0.01 M

Plug the values we get

E = E^o - (0.059 / n) xlog ([Al³⁺] / [Au⁺] )

   = +2.65 - (0.059 / 3 ) x log ( 0.02 / 0.01 )

   = +2.64 V

Therefore the EMF of the cell is +2.64 V