Determine the overall reaction and its standard cell potential at 25 °C for the reaction involving the galvanic cell in which cadmium metal is oxidized to 1 M cadmium(II) ion and a half-cell consisting of an aluminum electrode

in 1 M aluminum nitrate solution. Is the reaction spontaneous at standard conditions?

Short Answer

Expert verified

The overall reaction is \(3Cd(s) + 2A{l^{3 + }}(aq) \to 2Al(s) + 3C{d^{2 + }}(aq)\)

The standard cell potential at \({25\circ }{\rm{C}}\) is \( - 1.259\;{\rm{V}}\)

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

Step by step solution

01

Define standard cell potential

  • The shorthand notation of a cell reaction indicates phase boundary by a single vertical line (|) and salt bridge is denotes by a double vertical line . In this notation on the left side of salt bridge oxidation reaction which occurred at anode and on the right side of salt bridge; reduction reaction which occurred at cathode. The electrodes are written on the extreme corners; anode on extreme left and cathode on extreme right. The reactants in each half-cell are always first, followed by the products.
  • If the standard cell potential is positive, then the reaction isspontaneousand if the standard cell potential is negative, then the reaction isnon-spontaneous
02

Determine the standard cell potential

The cell reaction that involves cadmium electrode in \(1{\rm{MC}}{{\rm{d}}^{2 + }}\) ions and aluminium electrode in \(1{\rm{M}}\) aluminium nitrate therefore the cell notation is as follows:

\(Cd(s)\left| {C{d^{2 + }}(1M)} \right|\left| {A{l^{3 + }}(1M)} \right|Al(s)\)

In the given reaction cadmium is oxidized at anode and aluminium is reduced at cathode.

Anode reaction:

\(Cd(s) \to C{d^{2 + }}(aq) + 2{e^ - }\quad {E^o} = - 0.4030V\)

Cathode reaction:

\(A{l^{3 + }}(aq) + 3{e^ - } \to Al(s)\quad {E^o} = - 1.662\;{\rm{V}}\)

The overall or net cell reaction is as follows:

\(\begin{aligned}{l}3\left( {Cd(s) \to C{d^ + }(aq) + 2{e^ - }} \right)\\\frac{{2\left( {A{l^{3 + }}(aq) + 3{e^ - } \to Al(s)} \right)}}{{3Cd(s) + 2A{l^{3 + }}(aq) \to 2Al(s)}} + 3C{d^{2 + }}(aq)\end{aligned}\)Now calculate the standard cell potential at \({25\circ }{\rm{C}}\), using the following expression:

\(\begin{aligned}{l}E_{{\rm{cell }}}^o = E_{{\rm{cathode }}}^o - E_{{\rm{anode }}}^o\\E_{{\rm{cell }}}^o = - 1.662\;{\rm{V}} - ( - 0.4030\;{\rm{V}})\\ = - 1.259\;{\rm{V}}\end{aligned}\)Here the standard cell potential is negative, and then the reaction is non-spontaneous.

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Most popular questions from this chapter

Determine \({\bf{\Delta G}}\) and \({\bf{\Delta G}}^\circ \) for each of the reactions in the previous problem.

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

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

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

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

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