Question

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

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

Answer 1

Answer: (C)

The cell reaction may be split into two half reactions as

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

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

As $Zn ^{2+}$ ion has lower reduction potential than $Fe ^{2+}$ 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 $]$

$=-0.44-(-0.76)$

$=+0.35 \,V$