Question

The emf of a galvanic cell constituted with the electrodes $Z{n}^{2+}|Zn(-0.76V)$ and $F{e}^{2+}|Fe(-0.41V)$ is

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

Answer 1

Answer: (C)

The cell reaction may be split into two half reactions as

$Z{n}^{2+}+2e^{-}⟶Zn;E_{red}^{\circ }=-0.76,V$

$F{e}^{2+}+2e^{-}⟶Fe;E_{red}^{\circ }=-0.44V$

As $Z{n}^{2+}$ ion has lower reduction potential than $F{e}^{2+}$ ion.

Thus, oxidation takes place on zinc electrode standard EMF of the cell

$=[$ Standard reduction potential of the- the reduction half reaction $]-[$ Standard reduction potential ofthe oxidation half reaction $]$

$=-0.44-(-0.76)$

$=+0.35V$