Question

$ Ni/N{{i}^{2+}}[1.0M]||A{{u}^{3+}}[1.0M]/Au $ where $ E{}^\circ $ for $ N{{i}^{2+}}/Ni $ is $ -0.25\,V\,{{E}^{o}} $ for $ A{{u}^{3+}}/Au $ is 0.150V) What is the emf of the cell?

• A. + 0.4V
• B. $ -1.75V $
• C. + 1.25V
• D. + 1.75V

Answer 1

Answer: (A)

$ Ni/N{{i}^{2+}}[1.0M]||A{{u}^{3+}}[1.0M]/Au $ $ E_{cell}^{o}(A{{u}^{3+}}/Au)=0.15V $ $ E_{cell}^{o}(N{{i}^{2+}}/Ni)=-0.25V $ $ E_{cell}^{o}=E_{cathode}^{o}-E_{anode}^{o} $ $ =0.150-(-0.25) $ $ =+0.4V $