Question

calculate standard gibbs energy for cell ZnCl/Zn²+(aq) //Cu²+ (aq) /Au . E°Zn°/Zn=-.76v and Ecu°/cu =.34v . F= 96500c ​

Answer 1

Answer:

Explanation:

first find emf of cell

emf of cell=e(cathode)-e(anode)

=or defined as e(reduction potential  of cathode)-e(reduction potential of anode)

=.34-.76

=-.42

= -.42

gibbs free energy=-nfe

=.42*96500*2