Question

The emf of the cell, $Ni|N{i}^{2+}\left(1.0M\right)||A{g}^{+}\left(1.0M\right)|Ag$
$(E^{\circ }$ for $N{i}^{2+}/Ni=-0.25volt,E^{\circ }$ for $A{g}^{+}/Ag=0.80volt)$ is given by__________.

• A. $-0.25+0.80=0.55volt$
• B. $-0.25-(+0.80)=-1.05volt$
• C. $0+0.80-(-0.25)=+1.05volt$
• D. $-0.80-(-0.25)=-0.55volt$

Answer 1

The correct option is C $0+0.80-(-0.25)=+1.05volt$

In the given cell we have,

$Ni+2A{g}^{+}\to N{i}^{2+}+Ag$

so,

$E_{cell}=E_{cell}^0-\frac{0.0591}{2}log\frac{N{i}^{2+}}{(A{g}^{+}{)}^2}$

$E_{cell}=0.80-(-0.25)$- $\frac{0.0591}{2}log\left(1\right)$

$E_{cell}=0.80-(-0.25)=1.05volt$