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Chapter 17: Q17.4-29 E (page 960)

For the standard cell potentials given here, determine the \(\Delta {G\circ }\) for the cell in \({\rm{kJ}}\).

(a) \(0.000\;{\rm{V}},{\rm{n}} = 2\)

(b) \( + 0.434\;{\rm{V}},{\rm{n}} = 2\)

(c) \( - 2.439\;{\rm{V}},{\rm{n}} = 1\)

Short Answer

Expert verified
  1. The \(\Delta {G^o}\) for the cell is \(0.00\;{\rm{kJ}}\).
  2. The \(\Delta {G^o}\) for the cell is \( - 83.7\;{\rm{kJ}}\).
  3. The \(\Delta {G^o}\) for the cell is \( + 235.3\;{\rm{kJ}}\).

Step by step solution

01

Define standard cell potential

  • The electrical work of a cell at standard conditions with\(\Delta {G^{\rm{o}}}\), we get
  • \(\begin{aligned}{l}\Delta {G^o} = - nFE_{cell}^o\\F = {\rm{ Faraday costant }}(96,500{\rm{C}}/{\rm{mol}})\end{aligned}\)
  • For positive value of standard cell potential, the reaction is spontaneous and it is non-spontaneous for negative value of the standard cell potential.
02

a) Determine the standard cell potential

The \(\Delta {G^o}\) for the cell is \(0.00\;{\rm{kJ}}\).

Given that:

\(\begin{aligned}{l}E_{{\rm{cell }}}^{\rm{o}} = 0.00\;{\rm{V}}\\n = 2\end{aligned}\)

The electrical work of a cell at standard conditions with \(\Delta {G\circ }\) is calculated as follows:

\(\Delta {G^{\rm{o}}} = - nFE_{{\rm{cell}}}^{\rm{o}}\)

\(F = \)Faraday constant \((96,500{\rm{C}}/{\rm{mol}})\)

\(n = 2\).

The Gibbs free energy change when this quantity of reaction is converted is

\(\Delta {G^{\rm{o}}} = - \left( {\frac{{2{\rm{ mol }}{e^ - }}}{{{\rm{transferred }}}}} \right)\left( {\frac{{96,500{\rm{C}}}}{{{\rm{ mol }}{e^ - }}}} \right)\left( {\frac{{1\;{\rm{J}}}}{{1\;{\rm{V}} \times 1{\rm{C}}}}} \right)\left( {\frac{{1\;{\rm{kJ}}}}{{{{10}^3}\;{\rm{J}}}}} \right)(0.00\;{\rm{V}})\)

\(\Delta {G^{\rm{o}}} = 0.00\;{\rm{kJ}}\)

03

b) Determine the standard cell potential

The \(\Delta {G^o}\) for the cell is \( - 83.7\;{\rm{kJ}}\).

Given that:

\(\begin{aligned}{l}E_{{\rm{cell }}}^{\rm{o}} = + 0.434\;{\rm{V}}\\n = 2\end{aligned}\)

The electrical work of a cell at standard conditions with \(\Delta {G\circ }\) is calculated as follows:

\(\Delta {G^{\rm{o}}} = - nFE_{{\rm{cell }}}^{\rm{o}}\)

\(F = \)Faraday constant \((96,500{\rm{C}}/{\rm{mol}})\)

\(n = 2\). The Gibbs free energy change when this quantity of reaction is converted is

\(\begin{aligned}{l}\Delta {G^{\rm{o}}} = - \left( {\frac{{2{\rm{ mole }}{e^ - }}}{{{\rm{transferred }}}}} \right)\left( {\frac{{96,500{\rm{C}}}}{{{\rm{ mole }}{e^ - }}}} \right)\left( {\frac{{1\;{\rm{J}}}}{{1\;{\rm{V}} \times 1{\rm{C}}}}} \right)\left( {\frac{{1\;{\rm{kJ}}}}{{{{10}^3}\;{\rm{J}}}}} \right)( + 0.434\;{\rm{V}})\Delta {G^{\rm{o}}}\\ = - 83.7\;{\rm{kJ}}\end{aligned}\)

04

c) Determine the standard cell potential

The \(\Delta {G^o}\) for the cell is \( + 235.3\;{\rm{kJ}}\).

Given that:

\(\begin{aligned}{l}E_{{\rm{cell }}}^{\rm{o}} = - 2.439\;{\rm{V}}\\n = 1\end{aligned}\)

The electrical work of a cell at standard conditions with \(\Delta {G^{\rm{o}}}\) is calculated as follows:

\(\Delta {G^{\rm{o}}} = - nFE_{{\rm{cell }}}^{\rm{o}}\)

\(F = \)Faraday constant \((96,500{\rm{C}}/{\rm{mol}})\)

\(n = 1.\)The Gibbs free energy change when this quantity of reaction is converted is

\(\Delta {G^{\rm{o}}} = - \left( {\frac{{1\;{\rm{mole}}{{\rm{e}}^ - }}}{{{\rm{transferred }}}}} \right)\left( {\frac{{96,500{\rm{C}}}}{{{\rm{mol}}{{\rm{e}}^ - }}}} \right)\left( {\frac{{1\;{\rm{J}}}}{{1\;{\rm{V}} \times 1{\rm{C}}}}} \right)\left( {\frac{{1\;{\rm{kJ}}}}{{{{10}^3}\;{\rm{J}}}}} \right)( - 2.439\;{\rm{V}})\)

\(\Delta {G^{\rm{o}}} = + 235.3\;{\rm{kJ}}\)

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Most popular questions from this chapter

From the information provided, use cell notation to describe the following systems:

(a) In one half-cell, a solution of \({\rm{Pt}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\)forms Pt metal, while in the other half-cell, Cu metal goes into a \({\rm{Cu}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\)solution with all solute concentrations 1M.

(b) The cathode consists of a gold electrode in a \(0.55{\rm{MAu}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_3}\) solution and the anode is a magnesium electrode in \(0.75{\rm{MMg}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\) solution.

(c) One half-cell consists of a silver electrode in a \(1{\rm{MAgN}}{{\rm{O}}_3}\) solution, and in the other half-cell, a copper electrode in \(1M{\rm{Cu}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\) is oxidized.

Why must the charge balance in oxidation-reduction reactions?

Why would a sacrificial anode made of lithium metal be a bad choice despite its \({\bf{E}}_{{\bf{Li}}}^{\bf{^\circ }}{\bf{ + Li = - 3}}{\bf{.04\;V}}\), which appears to be able to protect all the other metals listed in the standard reduction potential table?

An active (metal) electrode was found to gain mass as the oxidation-reduction reaction was allowed to proceed. Was the electrode part of the anode or cathode? Explain.

Identify the species that was oxidized, the species that was reduced, the oxidizing agent, and the reducing agentin each of the reactions of the previous problem.

(a) \({\bf{S}}{{\bf{O}}_{\bf{3}}}^{{\bf{2 - }}}{\bf{(aq) + Cu(OH}}{{\bf{)}}_{\bf{2}}}{\bf{(s)}} \to {\bf{S}}{{\bf{O}}_{\bf{4}}}^{{\bf{2 - }}}{\bf{(aq) + Cu(OH)(s)}}\)

(b) \({{\bf{O}}_{\bf{2}}}{\bf{(g) + Mn(OH}}{{\bf{)}}_{\bf{2}}}{\bf{(s)}} \to {\bf{Mn}}{{\bf{O}}_{\bf{2}}}{\bf{(s)}}\)

(c) \({\bf{N}}{{\bf{O}}_{\bf{3}}}^{\bf{ - }}{\bf{(aq) + }}{{\bf{H}}_{\bf{2}}}{\bf{(g)}} \to {\bf{NO(g)}}\)

(d) \({\bf{Al(s) + Cr}}{{\bf{O}}_{\bf{4}}}^{{\bf{2 - }}}{\bf{(aq)}} \to {\bf{Al(OH}}{{\bf{)}}_{\bf{3}}}{\bf{(s) + Cr(OH}}{{\bf{)}}_{\bf{4}}}^{\bf{ - }}{\bf{(aq)}}\)
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