• # question_answer $Ni/N{{i}^{2+}}\,[1.0M]\,\,||A{{u}^{3+}}[1.0M]/Au$ (Where ${{E}^{o}}$ for $N{{i}^{2+}}/Ni$ is $-0.25\text{ }V,$ ${{E}^{o}}$ for $A{{u}^{3+}}/Au$is$0.150\text{ }V$) What is the emf of the cell? A)  $+1.25\,V$                     B)  $-1.75\,V$ nmn C)   $+1.75\text{ }V$                       D)  $+0.4\text{ }V$

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