The EMF of Cells

$\mathrm{E}°,$ for the half reactions are as

$$\begin{gathered} \mathrm{Zn}\to {\mathrm{Zn}}^{2+}+2{\mathrm{e}}^{-}\hspace{0.17em};\hspace{0.17em}\hspace{0.17em}\mathrm{E}°=0.76\hspace{0.17em}\mathrm{V} \\ \mathrm{Fe}\to {\mathrm{Fe}}^{2+}+2{\mathrm{e}}^{-}\hspace{0.17em};\hspace{0.17em}\hspace{0.17em}\mathrm{E}°=0.41\hspace{0.17em}\hspace{0.17em}\mathrm{V} \end{gathered}$$

The standard EMF for the cell reaction is:

${\mathrm{Fe}}^{2+}+\mathrm{Zn}\to {\mathrm{Zn}}^{2+}+\mathrm{Fe}$

Predict reaction of $1 \mathrm{M} \mathrm{HCl}$ with iron. Given, ${\mathrm{E}}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}^{°}=-0.44 \mathrm{V}$.

For the cell reaction $\mathrm{Zn} \left(\mathrm{s}\right)+{\mathrm{Mg}}^{2+}(1 \mathrm{M})={\mathrm{Zn}}^{2+}(1 \mathrm{M})+\mathrm{Mg},$ the $\mathrm{emf}$ has been found to be $1·60 \mathrm{V}.{\mathrm{E}}^0$ of the cell is

${\mathrm{E}}^0$ (red.) values of the half cells ${\mathrm{Mg}}^{2+}/\mathrm{Mg}$ and ${\mathrm{Cl}}_2/{\mathrm{Cl}}^{-}$ are respectively $-2·36 \mathrm{V}$ and $+1·36 \mathrm{V}.$ The ${\mathrm{E}}^0$ value of the cell $\mathrm{Mg}\left|{\mathrm{Mg}}^{2+}\right|\left|{\mathrm{Cl}}_2\right|{\mathrm{Cl}}^{-}$ is

The standard reduction potentials, ${\mathrm{E}}^0$, for the half-reactions are as

$\begin{array}{l}\mathrm{Zn}\to {\mathrm{Zn}}^{2+}+2{\mathrm{e}}^{-}; {\mathrm{E}}^0=+0.76 \mathrm{V}\\ \mathrm{Fe}\to {\mathrm{Fe}}^{2+}+2{\mathrm{e}}^{-}; {\mathrm{E}}^0=+0.41 \mathrm{V}\end{array}$

the $\mathrm{emf}$ for the cell reaction,

${\mathrm{Fe}}^{2+}+\mathrm{Zn}\to {\mathrm{Zn}}^{2+}+\mathrm{Fe}$ is

Does the physical size of a galvanic cell governs the potential that it will deliver? What does the size affect?

Construct galvanic cells from the following pairs of half cells and calculate their $\mathrm{EMF}$.

1) $\left(\mathrm{Pt}\right)$ ${\mathrm{H}}_2|\mathrm{HCl} (1 \mathrm{M})$ and $\mathrm{Pt} \left({\mathrm{Cl}}_2\right)|\mathrm{HCl} (1 \mathrm{M})$ $\left(1\mathrm{atm}\right)$

2) $\mathrm{Cu}\left|{\mathrm{Cu}}^{2+}\right(1 \mathrm{M})$ and ${\mathrm{Cl}}^{-}|{\mathrm{Cl}}_2 (\mathrm{Pt})$ $(1 \mathrm{M}) (1 \mathrm{atm})$

Use the $\mathrm{EMF}$ series.

$$\begin{gathered} {\mathrm{E}}_{{\mathrm{H}}^{+}|{\mathrm{H}}_2}^{°}=0 \\ {\mathrm{E}}_{{\mathrm{Cl}}_2|{\mathrm{Cl}}^{-}}^{°}=1.36 \mathrm{V} \end{gathered}$$

${\mathrm{E}}_{{\mathrm{Cu}}^{2+}|\mathrm{Cu}}^{°}=0.34 \mathrm{V}$