Question

$Ni/N{i}^{2+}[1.0M]||A{u}^{3+}[1.0M]/Au$ where $E{}^{\circ }$ for $N{i}^{2+}/Ni$ is $-0.25V{E}^o$ 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)

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