Question

EMF of cell; $Ni|N{i}^{2+}\left(1.0M\right)||A{u}^{2+}\left(1.0M\right)|Au$(Where $E^0$ for $N{i}^{2+}|Ni$ is $-0.25$V; $E^0$ for $A{u}^{+2}|Au$ is $1.50$V) is:

• A. $+1.25$V
• B. $-1.75$V
• C. $+1.75$V
• D. $+4.0$V

Answer 1

Answer: (C)

$+1.75$V

Using Nearest equation :-

$\begin{array}{c}{E}_{cell}=E{}_{cell}^{\circ }-\frac{0.0593}{nRT}\log \frac{\left[N{i}^{2+}\right]}{\left[A{u}^{2+}\right]}\\ E{}_{cell}^{\circ }=E{}_{cathode}^{\circ }-E{}_{anode}^{\circ }\Rightarrow 1.50V-\left(-0.25V\right)\Rightarrow 1.75V\\ E_{cell}=E{}_{cell}^{\circ }-\frac{0.0593}{nRT}\log \frac{1}{1}\Rightarrow E{}_{cell}^{\circ }-0\\ E_{cell}=E{}_{cell}^{\circ }=1.75V\end{array}$

Option C is correct.