For each reaction listed, determine its standard cell potential at \({25\circ }{\rm{C}}\) and whether the reaction is spontaneous at standard conditions.

(a)\({\mathop{\rm Mn}\nolimits} (s) + {\rm{N}}{{\rm{i}}^{2 + }}(aq) \to {{\mathop{\rm Mn}\nolimits} ^{2 + }}(aq) + {\rm{Ni}}(s)\)

(b)\(3{\rm{C}}{{\rm{u}}^{2 + }}(aq) + 2{\rm{Al}}(s) \to 2{\rm{A}}{{\rm{l}}^{3 + }}(aq) + 3{\rm{Cu}}(s)\)

(c)\({\rm{Na}}(s) + {\rm{LiN}}{{\rm{O}}_3}(aq) \to {\rm{NaN}}{{\rm{O}}_3}(aq) + {\rm{Li}}(s)\)

(d) \({\rm{Ca}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}(aq) + {\rm{Ba}}(s) \to {\rm{Ba}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}(aq) + {\rm{Ca}}(s)\)

For positive value of standard cell potential, the reaction isspontaneousand it isnon-spontaneousfor negative value of the standard cell potential.

The standard cell potential at \({25\circ }{\rm{C}}\) is \( + 1.187\;{\rm{V}}\)

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

Given:

\(Mn(s) + N{i^{2 + }}(aq) \to M{n^{2 + }}(aq) + Ni(s)\)

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

Anode reaction:

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

Cathode reaction:

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

Now calculate the standard cell potential at \({25\circ }{\rm{C}}\), using the following expression:

\(\begin{aligned}E_{{\rm{cell }}}^o &= E_{{\rm{cathode }}}^o - E_{{\rm{anode }}}^o\\E_{{\rm{cell }}}^o &= - 1.185\;{\rm{V}} - ( - 2.372\;{\rm{V}})\\ &= + 1.187\;{\rm{V}}\end{aligned}\)

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 \({25\circ }{\rm{C}}\), using the following expression:

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

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

Given:

\({\rm{Na}}(s) + {\rm{LiN}}{{\rm{O}}_3}(aq) \to {\rm{NaN}}{{\rm{O}}_3}(aq) + {\rm{Li}}({\rm{s}})\)

In the given reaction sodium is oxidized at anode and lithium is reduced at cathode

Anode reaction:

\(Na(s) \to N{a^ + }(aq) + {e^ - }\quad {E^o} = - 2.71\;{\rm{V}}\)

Cathode reaction:

\({\rm{L}}{{\rm{i}}^ + }(aq) + {e^ - } \to {\rm{Li}}(s)\quad {E^o} = - 3.04\;{\rm{V}}\)

Now, calculate the standard cell potential at \({25\circ }{\rm{C}}\), using the following expression:

\(\begin{aligned}E_{{\rm{cell }}}^o &= E_{{\rm{cathode }}}^o - E_{{\rm{anode }}}^o\\E_{{\rm{cell }}}^o &= - 3.04\;{\rm{V}} - ( - 2.71\;{\rm{V}})\\ &= - 0.33\;{\rm{V}}\end{aligned}\)

Here the standard cell potential is negative, and then the reaction is non-spontaneous.

Given:

\({\rm{Ca}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}(aq) + {\rm{Ba}}({\rm{s}}) \to {\rm{Ba}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}(aq) + {\rm{Ca}}({\rm{s}})\)

In the given reaction barium is oxidized at anode and calcium is reduced at cathode

Anode reaction:

\(Ba(S) \to B{a^{2 + }}(aq) + 2{e^ - }\quad {E^o} = - 2.912V\)

Cathode reaction:

Now calculate the standard cell potential at \({25\circ }{\rm{C}}\), using the following expression:

\(\begin{aligned}E_{{\rm{cell }}}^o &= E_{{\rm{cathode }}}^o - E_{{\rm{anode }}}^o\\E_{{\rm{cell }}}^o &= - 2.868 - ( - 2.912V)\\ & = + 0.044V\end{aligned}\)

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