question_answer

The standard electrode potential of the half cells is given below:

\[Z{{n}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Zn;E=-7.62\,V,\]

\[F{{e}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Fe;E=-7.81\,V,\]

The emf of the cell \[F{{e}^{2+}}+Zn\xrightarrow{{}}Z{{n}^{2+}}+Fe\]is

A)  1.54 V                         

B)  -1.54 V

C)  -0.19 V                        

D)  +0.19 V

Correct Answer: C

Solution :

\[EMF={{E}_{cathode}}-Eanode=-7.81-(-7.62)\] \[=EMF=-0.19\,V.\]