Question

The standard reduction potentials $E^{⊝}$ for the half reactions are follows :

$Zn\to Z{n}^{2+}+2e^{-};E^{⊝}=+0.76V$
$Fe\to F{e}^{2+}+2e^{-};E^{⊝}=0.41V$

The EMF for the cell reaction $F{e}^{2+}Zn\to Z{n}^{2+}+Fe$ is:

• A. $-0.35V$
• B. $+0.35V$
• C. $+1.17V$
• D. $-1.17V$

Answer 1

Answer: (B)

$+0.35V$

$F{e}^{2+}+Zn\to Z{n}^{2+}+Fe$
$Zn\to Z{n}^{2+}+2e^{-};E^{⊝}=+0.76V$
$Fe\to F{e}^{2+}+2e^{-};E^{⊝}=+0.41V$
These are oxidation potentials.
Reduction potentials are equal and opposite.
Fe forms cathode and Zn forms anode.
$E_{cell}^{⊝}=(E_{red}^{⊝}{)}_c+(E_{oxid}^{⊝}{)}_a$
= (-0.41 + 0.76) V
= 0.35 V