Question

The emf of the cell, $Ni|{Ni}^{2+}\left(1.0M\right)\parallel {Ag}^{+}\left(1.0M\right)|Ag[E^{o}for{Ni}^{2+}/Ni=-0.25volt,E^{o}for{Ag}^{+}/Ag=0.80volt]$ is given by :${[E^{o}forAg}^{+}/Ag=0.80volt]$

• 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 $0+0.80-(-0.25)=+1.05volt$

$Ni⟶{Ni}^{2+}+{2e}^{-}2\{{Ag}^{2+}+e^{-}⟶Ag\}\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_Ni+2{Ag}^{2+}⟶{Ni}^{2+}+2Ag{E}_{cell}^o=0.08-(-0.25)=1.05VE=E_{cell}^o-\frac{0.0591}{2}log\frac{\left[{Ni}^{2+}\right]}{{\left[{Ag}^{+}\right]}^2}{E}_{cell}=1.05-\frac{0.0591}{2}\times 0=1.05$