Question

Sketch the galvanic cells based on the following overall 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. Standard reduction potentials are found in Table 11.1.

$\begin{array}{l}\mathbf{a}\mathbf{.}{\mathbf{Cr}}^{\mathbf{3}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{CI}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\rightleftarrows }{\mathbf{Cr}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{7}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{CI}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\\ \mathbf{b}\mathbf{.}{\mathbf{Cu}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{Mg}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\rightleftarrows }{\mathbf{Mg}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{Cu}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\\ \mathbf{c}\mathbf{.}{{\mathbf{IO}}_{\mathbf{3}}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{Fe}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\rightleftarrows }{\mathbf{Fe}}^{\mathbf{3}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{I}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\\ \mathbf{d}\mathbf{.}\mathbf{Zn}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathbf{Ag}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\rightleftarrows }{\mathbf{Zn}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{Ag}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\end{array}$

Short answer

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

a.0.03V

b.2.17V

c.0.43V

d 1.56V

Step-by-step solution

Definition of galvanic cell

A galvanic cell is an electro chemical cell that drives electrical energy from chemical reactionstaking place within a particular cell.In a galvanic cell the oxidation and the reduction half cells are separated by a connecting wire so that electrons can flow through the wire.

Explanation of a.Cr3+(aq)+CI2(g)⇄Cr2O72-(aq)+CI-(aq)

$\begin{array}{l}{\mathrm{E}}^0={{\mathrm{E}}^0}_{\mathrm{cathode}}-{{\mathrm{E}}^0}_{\mathrm{anode}}\\ {\mathrm{E}}^0=1.36-1.33\\ {\mathrm{E}}^0=0.03\mathrm{V}\end{array}$

Explanation of b.Cu2+(aq)+Mg(s)⇄Mg2+(aq)+Cu(s)

$\begin{array}{l}{\mathrm{E}}^0={{\mathrm{E}}^0}_{\mathrm{cathode}}-{{\mathrm{E}}^0}_{\mathrm{anode}}\\ {\mathrm{E}}^0=0.34-2.37\\ {\mathrm{E}}^0=2.71\mathrm{V}\end{array}$

Explanation of c.IO3-(aq)+Fe2+(aq)⇄Fe3+(aq)+I2(s)

$\begin{array}{l}{\mathrm{E}}^0={{\mathrm{E}}^0}_{\mathrm{cathode}}-{{\mathrm{E}}^0}_{\mathrm{anode}}\\ {\mathrm{E}}^0=1.20-0.77\\ {\mathrm{E}}^0=0.43\mathrm{V}\end{array}$

Explanation of d.Zn(s)+Ag+(aq)⇄Zn2+(aq)+Ag(s)

$\begin{array}{l}{\mathrm{E}}^0={{\mathrm{E}}^0}_{\mathrm{cathode}}-{{\mathrm{E}}^0}_{\mathrm{anode}}\\ {\mathrm{E}}^0=0.80-0.76\\ {\mathrm{E}}^0=1.56\mathrm{V}\end{array}$