Question

Calculate the open circuit potential of a Daniell cell (at 25 oC and 1 bar) with ZnSO4(aq, 0.01 mol/kg ) on the left hand side and CuSO4(aq, 0.002 mol/kg) on the right hand side if the ionic activity coefficients of Cu2+(aq) and Zn2+(aq) are, respectively, 0.658 and 0.392.

Answer 1

Cu is cathode & Zn is anode

Cu²⁺(aq)+2e^--->Cu(s)

Zn(s)-->Zn²⁺(aq)+2e^-

note that n=2 for this cell

activities for solid materials is one

=

& =1

therefore

on substing all values

E= 1.1-8.314*298/(2*96 485.3329)ln(0.392*0.01/(0.658*0.002)=1.088V

so open circuit potential is 1.088V