question_answer

The standard reduction potential,\[E{}^\circ \]for the half-reactions are as \[Zn\,\,\,\,Z{{n}^{2+}}+2{{e}^{-}};\] \[E{}^\circ =+\,0.76\,V\] \[Fe\,\,\,\,F{{e}^{2+}}+2{{e}^{-}};\]\[E{}^\circ =+\,0.41\,V\] The \[E_{cell}^{\text{o}}\] for the cell formed by these two electrodes is

A)  - 0.35 V               

B)  - 1.17 V

C)  + 0.35 V              

D)  + 1.17 V

Correct Answer: C

Solution :

                In this reaction, \[EMF={{E}_{cathode}}-{{E}_{anode}}\]            \[=-\,0.41-(-0.76)=+\,0.35\,V\]