For each reaction listed, determine its standard cell potential at 25^oC and whether the reaction is spontaneous at standard conditions.

(a) \({\rm{Mg}}(s) + {\rm{N}}{{\rm{i}}^{2 + }}(aq) \to {\rm{M}}{{\rm{g}}^{2 + }}(aq) + {\rm{Ni}}(s)\)

(b) \(2{\rm{A}}{{\rm{g}}^ + }(aq) + {\rm{Cu}}(s) \to {\rm{C}}{{\rm{u}}^{2 + }}(aq) + 2{\rm{Ag}}(s)\)

(c) \({\rm{Mn}}(s) + {\rm{Sn}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}(aq) \to {\mathop{\rm Mn}\nolimits} {\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}(aq) + {\mathop{\rm Sn}\nolimits} (s)\)

(d) \(3{\rm{Fe}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}(aq) + {\rm{Au}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_3}(aq) \to 3{\rm{Fe}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_3}(aq) + {\rm{Au}}(s)\)

• Avoltaic or a galvanic cellis an electrochemical cell which derives the electrical energy from the spontaneous redox chemical reactions which occur inside the cell.
• A galvanic cell usually contains two different metals which are dipped in an electrolytic solution or of an individual half-cells having different metals and their ions present in the solution, joined by a salt-bridge or divided by a porous membrane.

Given:

\(Mg(s) + N{i^{2 + }}(aq) \to M{g^{2 + }}(aq) + Ni(s)\)

In the given reaction Mg is oxidized at anode and Ni is reduced at cathode

Anode reaction:

\(Mg(s) \to M{g^{2 + }}(aq) + 2{e^ - }\quad {E^o} = - 2.372\;{\rm{V}}\)

Cathode reaction:

\(N{i^{2 + }}(aq) + 2{e^ - } \to Ni(s)\quad {E^o} = - 0.257V\)

Now calculate the standard cell potential at 25^oC using the following expression:

\(\begin{array}{l}E_{{\rm{cell }}}^o = E_{{\rm{cathode }}}^o - E_{{\rm{anode }}}^o\\E_{{\rm{cell }}}^o = - 0.257\;{\rm{V}} - ( - 2.372\;{\rm{V}})\\ = + 2.115\;{\rm{V}}\end{array}\)

Here the standard cell potential is positive, and then the reaction is spontaneous.

Given:

\(Cu(s) + 2A{g^ + }(aq) \to C{u^{2 + }}(aq) + 2Ag(s)\)

In the given reaction copper is oxidized at anode and silver is reduced at cathode

Anode reaction:

\(Cu(s) \to {\rm{C}}{{\rm{u}}^{2 + }}(aq) + 2{e^ - }\quad {E^o} = + 0.34\;{\rm{V}}\)

Cathode reaction:

\(2{\rm{A}}{{\rm{g}}^ + }(aq) + 2{e^ - } \to 2{\rm{Ag}}(s)\quad {E^o} = + 0.7996\;{\rm{V}}\)

Now calculate the standard cell potential at 25^oC using the following expression:

\(\begin{array}{l}E_{{\rm{cell }}}^o = E_{{\rm{cathode }}}^o - E_{{\rm{anode }}}^o\\E_{{\rm{cell }}}^o = + 0.7996\;{\rm{V}} - (0.34\;{\rm{V}})\\ = + 0.4596\;{\rm{V}}\end{array}\)

Here the standard cell potential is positive, and then the reaction is spontaneous.

The standard cell potential at 25^oC is \( + 1.0475\;{\rm{V}}\)

The standard cell potential is positive, and then the reaction is spontaneous.

Explanation of Solution

Given:

\(Mn(s) + {\mathop{\rm Sn}\nolimits} {\left( {{N_3}} \right)_2}(aq) \to Mn{\left( {N{O_3}} \right)_2}(aq) + Sn(s)\)

In the given reaction manganese is oxidized at anode and tin is reduced at cathode Anode reaction:

\(Mn(s) \to M{n^{2 + }}(aq) + 2{e^ - }\quad {E^o} = - 1.185\;{\rm{V}}\)

Cathode reaction:

\({{\mathop{\rm Sn}\nolimits} ^{2 + }}(aq) + 2{e^ - } \to {\mathop{\rm Sn}\nolimits} (s)\quad {E^o} = - 0.1375\;{\rm{V}}\)

Now calculate the standard cell potential at 25^oC using the following expression:

\(\begin{array}{l}E_{{\rm{cell }}}^o = E_{{\rm{cathode }}}^o - E_{{\rm{anode }}}^o\\E_{{\rm{cell }}}^o = - 0.1375\;{\rm{V}} - ( - 1.185\;{\rm{V}})\\ = + 1.0475\;{\rm{V}}\end{array}\)

Here the standard cell potential is positive, and then the reaction is spontaneous.

Given:

\(3{\rm{Fe}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}(aq) + {\rm{Au}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_3}(s) \to 3{\rm{Fe}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_3}(aq) + {\rm{Au}}({\rm{s}})\)

In the given reaction iron is oxidized from \( + 2\) to \( + 3\) at anode and gold is reduced at cathode

Anode reaction:

\({\rm{F}}{{\rm{e}}^{2 + }}({\rm{aq}}) \to {\rm{F}}{{\rm{e}}^{3 + }}({\rm{aq}}) + {{\rm{e}}^ - }\quad {E^o} = + 0.771V\)

Cathode reaction:

\({\rm{A}}{{\rm{u}}^{3 + }}({\rm{aq}}) + 3{{\rm{e}}^ - } \to {\rm{Au}}({\rm{s}})\quad {E^o} = + 1.498V\)

Now calculate the standard cell potential at 25^oC using the following expression:

\(\begin{array}{l}E_{{\rm{cell }}}^o = E_{{\rm{cathode }}}^o - E_{{\rm{anode }}}^o\\E_{{\rm{cell }}}^o = + 1.498\;{\rm{V}} - (0.771\;{\rm{V}})\\ = + 0.727\;{\rm{V}}\end{array}\)

Here the standard cell potential is positive, and then the reaction is spontaneous.