Question

Calculate the potential at 25^o C for the following cell reaction.

a) Cu (s)+ 2Ag (aq) --> Cu²⁺ (aq) + 2Ag (s)

b) 3Cu²⁺ (aq) +2Al (s) --> 3Cu (s) +2Al³⁺

Answer 1

a) Cu (s)+ 2Ag (aq) --> Cu²⁺ (aq) + 2Ag (s)

At anode ( oxidation) : Cu (s) --> Cu²⁺ (aq) +2e^-   :   E^o_Cu2+/Cu = +0.34 V

At cathode ( reduction) : Ag⁺ (aq)+e^- --> Ag (s)    : E^o_Ag/Ag+ = +0.80 V

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

                                                      = E^o_Ag/Ag+ - E^o_Cu2+/Cu

                                                      = +0.80 - (+0.34) V

                                                      = +0.46 V

Therefore the standard potential of the cell is +0.46 V

b) 3Cu²⁺ (aq) +2Al (s) --> 3Cu (s) +2Al³⁺

At anode ( oxidation) : Al (s) --> Al³⁺+3e^-   :   E^o_Al3+/Al = -1.66 V

At cathode ( reduction) : Cu²⁺ (aq) +2e^- --> Cu (s) : E^o_Cu2+/Cu = +0.34 V

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

                                                      =   E^o_Cu2+/Cu - E^o_Al3+/Al

                                                      = +0.34 - (-1.66) V

                                                      = +2.00 V

Therefore the standard potential of the cell is +2.00 V