Question

Sketch the galvanic cells based on the following half-reactions. Calculate, show the direction of electron flow and the direction of ion migration through the salt bridge, identify the cathode and anode, and give the overall balanced equation. Assume that all concentrations are 1.0 M and that all partial pressures are 1.0 atm.

$\begin{array}{l}\mathbf{a}\mathbf{.}{\mathbf{CI}}_{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathbf{e}}^{\mathbf{-}}\mathbf{\to }\mathbf{2}{\mathbf{CI}}^{\mathbf{-}}\mathbf{,}{\mathbf{E}}^{\mathbf{0}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{36}\mathbf{V}\\ {\mathbf{Br}}_{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathbf{e}}^{\mathbf{-}}\mathbf{\to }\mathbf{2}{\mathbf{Br}}^{\mathbf{-}}\mathbf{,}{\mathbf{E}}^{\mathbf{0}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{09}\mathbf{V}\\ \mathbf{b}\mathbf{.}{\mathbf{MnO}}_{\mathbf{4}}\mathbf{+}\mathbf{8}{\mathbf{H}}^{\mathbf{+}}\mathbf{+}\mathbf{5}{\mathbf{e}}^{\mathbf{-}}\mathbf{\to }{\mathbf{Mn}}^{\mathbf{2}\mathbf{+}}\mathbf{+}\mathbf{4}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{,}{\mathbf{E}}^{\mathbf{0}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{51}\mathbf{V}\\ {{\mathbf{IO}}_{\mathbf{4}}}^{\mathbf{-}}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{+}\mathbf{2}{\mathbf{e}}^{\mathbf{-}}\mathbf{\to }{{\mathbf{IO}}_{\mathbf{3}}}^{\mathbf{-}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{,}{\mathbf{E}}^{\mathbf{0}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{60}\mathbf{V}\\ \mathbf{c}\mathbf{.}{\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{+}\mathbf{2}{\mathbf{e}}^{\mathbf{-}}\mathbf{\to }\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{,}{\mathbf{E}}^{\mathbf{0}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{78}\mathbf{V}\\ {\mathbf{O}}_{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{+}\mathbf{2}{\mathbf{e}}^{\mathbf{-}}\mathbf{\to }{\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\mathbf{,}{\mathbf{E}}^{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{68}\mathbf{V}\\ \mathbf{d}\mathbf{.}{\mathbf{Mn}}^{\mathbf{2}\mathbf{+}}\mathbf{+}\mathbf{2}{\mathbf{e}}^{\mathbf{-}}\mathbf{\to }\mathbf{Mn}\mathbf{,}{\mathbf{E}}^{\mathbf{0}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{18}\mathbf{V}\\ {\mathbf{Fe}}^{\mathbf{3}\mathbf{+}}\mathbf{+}\mathbf{3}{\mathbf{e}}^{\mathbf{-}}\mathbf{\to }\mathbf{Fe}\mathbf{,}{\mathbf{E}}^{\mathbf{0}}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{036}\mathbf{V}\end{array}$

Short answer

The values of${{\mathrm{E}}^0}_{\mathrm{cell}}$ are,

a.0.27V

b.0.09V

c.1.10V

d.1.14V

Step-by-step solution

Definition of Energies

In galvanic cell chemical energy converted into electrical energy.

At anode oxidation takes place which means loss of electrons.

At cathode reduction takes place which means gain of electrons

Explanation of CI2+2e-→2CI-,E0=1.36V;Br2+2e-→2Br-,E0=1.09V

$\begin{array}{l}{\mathrm{E}}^0={{\mathrm{E}}^0}_{\mathrm{cathode}}-{{\mathrm{E}}^0}_{\mathrm{anode}}\\ {\mathrm{E}}^0=1.36-(-1.09)\\ {\mathrm{E}}^0=0.27\mathrm{V}\end{array}$

Explanation of b.MnO4+8H++5e-→Mn2++4H2O,E0=1.51V;IO4-+2H++2e-→IO3-+H2O,E0=1.60V

$\begin{array}{l}{\mathrm{E}}^0={{\mathrm{E}}^0}_{\mathrm{cathode}}-{{\mathrm{E}}^0}_{\mathrm{anode}}\\ {\mathrm{E}}^0=1.60-1.51\\ {\mathrm{E}}^0=0.09\mathrm{V}\end{array}$

Explanation of c.H2O2+2H++2e-→2H2O,E0=1.78V;O2+2H++2e-→H2O2,E0=0.68V

$\begin{array}{l}{\mathrm{E}}^0={{\mathrm{E}}^0}_{\mathrm{cathode}}-{{\mathrm{E}}^0}_{\mathrm{anode}}\\ {\mathrm{E}}^0=1.78-0.68\\ {\mathrm{E}}^0=1.10\mathrm{V}\end{array}$

d.Mn2++2e-→Mn,E0=-1.18V;Fe3++3e-→Fe,E0=-0.036V

$\begin{array}{l}{E}^0={E^0}_{cathode}-{E^0}_{anode}\\ E^0=-0.036-(-1.18)\\ E^0=1.14V\end{array}$