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Chapter 17: Q17.4-30 E (page 960)

For the \(\Delta {G\circ }\) values given here, determine the standard cell potential for the cell.

(a) \(12\;{\rm{kJ}}/{\rm{mol}},{\rm{n}} = 3\)

(b) \( - 45\;{\rm{kJ}}/{\rm{mol}},{\rm{n}} = 1\)

Short Answer

Expert verified
  1. The standard cell potential from given \(\Delta {G^o}\) for the cell is \( - 0.0415\;{\rm{V}}\)
  2. The standard cell potential from given \(\Delta {G^o}\) for the cell is \( + 0.466\;{\rm{V}}\)

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
  • \(\begin{aligned}{}\Delta {G^o} = - nFE_{cell}^o\\F = {\rm{ Faraday costant }}(96,500{\rm{C}}/{\rm{mol}})\end{aligned}\)
  • 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 standard cell potential from given \(\Delta {G^o}\) for the cell is \( - 0.0415\;{\rm{V}}\)

Given that:

\(\begin{aligned}{}\Delta {G^o} = 12\;{\rm{kJ}}/{\rm{mol}}\\n = 3\end{aligned}\)

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

\(\begin{aligned}{}\Delta {G^o} = - nFE_{cell}^o\\F = {\rm{ Faraday costant }}(96,500{\rm{C}}/{\rm{mol}})\end{aligned}\)

The standard cell potential is calculated as follows:

\(\begin{aligned}{}12\;{\rm{kJ}}/{\rm{ mole }} = - \left( {3{\rm{ mole}}{{\rm{ }}^ - }{\rm{transferred }}} \right)\left( {\frac{{96,500{\rm{C}}}}{{{\rm{ mole}}{{\rm{ }}^ - }}}} \right)\left( {\frac{{1\;{\rm{J}}}}{{1V \times 1{\rm{C}}}}} \right)\left( {\frac{{1\;{\rm{kJ}}}}{{{{10}^3}\;{\rm{J}}}}} \right)\left( {E_{{\rm{cell }}}^o{\rm{ inV }}} \right)\\E_{{\rm{cell }}}^o = - 0.0415\;{\rm{V}}\end{aligned}\)

03

b) Determine the standard cell potential

The standard cell potential from given \(\Delta {G^o}\) for the cell is \( + 0.466\;{\rm{V}}\)

Explanation of Solution

\(\begin{aligned}{}\Delta {G^o} = - 45\;{\rm{kJ}}/{\rm{mol}}\\n = 1\end{aligned}\)

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

\(\begin{aligned}{}\Delta {G^o} = - nFE_{cell}^o\\F = {\rm{ Faraday costant }}(96,500{\rm{C}}/{\rm{mol}})\end{aligned}\)

The standard cell potential is calculated as follows:

\(\begin{aligned}{} - 45\;{\rm{kJ}}/{\rm{ mole }} = - \left( {1{\rm{ mole}}{{\rm{ }}^ - }{\rm{transferred }}} \right)\left( {\frac{{96,500{\rm{C}}}}{{{\rm{ mole}}{{\rm{ }}^ - }}}} \right)\left( {\frac{{1\;{\rm{J}}}}{{1V \times 1{\rm{C}}}}} \right)\left( {\frac{{1\;{\rm{kJ}}}}{{{{10}^3}\;{\rm{J}}}}} \right)\left( {E_{{\rm{cell }}}^o{\rm{ inV }}} \right)\\E_{{\rm{cell }}}^o = + 0.466\;{\rm{V}}\end{aligned}\)

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Most popular questions from this chapter

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

Consider the following metals: Ag, Au, \(Mg, Ni,\)\(and\)\(Zn\). Which of these metals could be used as a sacrificial anode in the cathodic protection of an underground steel storage tank? Steel is mostly iron, so use \( - 0.447\;{\rm{V}}\) as the standard reduction potential for steel.

Why is it not possible for hydroxide ion \(\left( {{\bf{O}}{{\bf{H}}^ - }} \right)\)to appear in either of the half-reactions or the overall equation when balancing oxidation-reduction reactions in acidic solution?

Use cell notation to describe the galvanic cell where copper(II) ions are reduced to copper metal and zinc metal is oxidized to zinc ions.

Given the following cell notations, determine the species oxidized, species reduced, and the oxidizing agent and reducing agent, without writing the balanced reactions

a. \({\rm{Mg}}(s)\left| {{\rm{M}}{{\rm{g}}^{2 + }}(aq) || {\rm{C}}{{\rm{u}}^{2 + }}(aq)} \right|{\rm{Cu}}(s)\)

b.\({\rm{Ni}}(s)\left| {{\rm{N}}{{\rm{i}}^{2 + }}(aq) || {\rm{A}}{{\rm{g}}^ + }(aq)} \right|{\rm{Ag}}(s)\)
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