MCQ
The standard reduction potential for the half reactions are as
$Zn \rightarrow Zn ^{2+}+2 e ^{-} E ^{\circ}=+0.76 V$
$Fe \rightarrow Fe ^{2+}+2 e ^{-} E ^{\circ}=+041 V .$
So for cell reaction $F ^{2+}+ Zn \rightarrow Zn ^{2+}+ Fe$ is
$Zn \rightarrow Zn ^{2+}+2 e ^{-} E ^{\circ}=+0.76 V$
$Fe \rightarrow Fe ^{2+}+2 e ^{-} E ^{\circ}=+041 V .$
So for cell reaction $F ^{2+}+ Zn \rightarrow Zn ^{2+}+ Fe$ is
- A$– 0.35V$
- ✓$+0.35V$
- C$+1.17V$
- D$– 1.117V$