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

The cell reaction may be split into two half reactions as
\(\begin{aligned}
&\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn} ; \quad E_{\mathrm{red}}^{\circ}=-0.76 \mathrm{~V} \\
&\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe} ; \quad E_{\mathrm{red}}^{\circ}=-0.44 \mathrm{~V}
\end{aligned}\)
As \(\mathrm{Zn}^{2+}\) ion has lower reduction potential than \(\mathrm{Fe}^{2+}\) ion. Thus, oxidation takes place on zinc electrode standard EMF of the cell
\(=\left[\begin{array}{l}\text { Standard reduction potential of the } \\ \text { the reduction half reaction }\end{array}\right]-\)
\(\left[\begin{array}{l}\text { Standard reduction potential of } \\ \text { the oxidation half reaction }\end{array}\right]\)
\(=-0.44-(-0.76)\)
\(=+0.35 \mathrm{~V}\)