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Chapter 17: Q17.3-27 E (page 959)

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

Aluminium\(\left( {{\bf{E}}_{{\bf{A}}{{\bf{l}}^{{\bf{3 + }}}}{\bf{/Al}}}^{\bf{^\circ }}{\bf{ = - 2}}{\bf{.07\;V}}} \right)\) is more easily oxidized than iron \(\left( {{\bf{E}}_{{\bf{F}}{{\bf{e}}^{\bf{3}}}}^{\bf{^\circ }}{\bf{/F}}{{\bf{e}}^{\bf{ - }}}{\bf{ = - 0}}{\bf{.477\;V}}} \right){\bf{,}}\) and yet when both are exposed to the environment, untreated aluminium has very good corrosion resistance while the corrosion resistance of untreated iron is poor. Explain this observation.

Determine the overall reaction and its standard cell potential at 25 °C for the reaction involving the galvanic cell made from a half-cell consisting of a silver electrode in 1 M silver nitrate solution and a half-cell consisting of azinc electrode in 1 M zinc nitrate. Is the reaction spontaneous at standard conditions?

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

Given the following pairs of balanced half-reactions, determine the balanced reaction for each pair of half reactions in an acidic solution.

(a) \({\bf{Ca}} \to {\bf{C}}{{\bf{a}}^{{\bf{2 + }}}}{\bf{ + 2}}{{\bf{e}}^{\bf{ - }}}{\bf{,}}\quad {{\bf{F}}_{\bf{2}}}{\bf{ + 2}}{{\bf{e}}^{\bf{ - }}} \to {\bf{2\;}}{{\bf{F}}^{\bf{ - }}}\)

(b) \({\bf{Li}} \to {\bf{L}}{{\bf{i}}^{\bf{ + }}}{\bf{ + }}{{\bf{e}}^{\bf{ - }}}{\bf{,}}\quad {\bf{C}}{{\bf{l}}_{\bf{2}}}{\bf{ + 2}}{{\bf{e}}^{\bf{ - }}} \to {\bf{2C}}{{\bf{l}}^{\bf{ - }}}\)

(c) \({\bf{Fe}} \to {\bf{F}}{{\bf{e}}^{{\bf{3 + }}}}{\bf{ + 3}}{{\bf{e}}^{\bf{ - }}}{\bf{,}}\quad {\bf{B}}{{\bf{r}}_{\bf{2}}}{\bf{ + 2}}{{\bf{e}}^{\bf{ - }}} \to {\bf{2B}}{{\bf{r}}^{\bf{ - }}}\)

(d) \({\bf{Ag}} \to {\bf{A}}{{\bf{g}}^{\bf{ + }}}{\bf{ + }}{{\bf{e}}^{\bf{ - }}}{\bf{,}}\quad {\bf{MnO}}_{\bf{4}}^{\bf{ - }}{\bf{ + 4}}{{\bf{H}}^{\bf{ + }}}{\bf{ + 3}}{{\bf{e}}^{\bf{ - }}} \to {\bf{Mn}}{{\bf{O}}_{\bf{2}}}{\bf{ + 2}}{{\bf{H}}_{\bf{2}}}{\bf{O}}\)

Identify the species that undergoes oxidation, the species that undergoes reduction, the oxidizing agent, and thereducing agent in each of the reactions of the previous problem.

\(\begin{array}{l}{\bf{(a) H_2O_2 + S}}{{\bf{n}}^{{\bf{2 + }}}} \to {\bf{H_2O + S}}{{\bf{n}}^{{\bf{4 + }}}}\\{\bf{(b) PbO_2 + Hg}} \to {\bf{Hg}}{{\bf{2}}^{{\bf{2 + }}}}{\bf{ + P}}{{\bf{b}}^{{\bf{2 + }}}}\\{\bf{(c) Al + Cr_2O}}{{\bf{7}}^{{\bf{2 - }}}} \to {\bf{A}}{{\bf{l}}^{{\bf{3 + }}}}{\bf{ + C}}{{\bf{r}}^{{\bf{3 + }}}}\end{array}\)
See all solutions

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