Calculating the Free-Energy Change from Electrode Potentials Using standard electrode potentials, calculate the standard free-energy change at 25°C for the reaction Zn(s) + 2Ag^+(aq) → Zn^2+(aq) + 2Ag(s) PROBLEM STRATEGY Substitute into ΔG° = -nFE° cell. Use a table of standard electrode potentials

Question

Calculating the Free-Energy Change from Electrode Potentials

Using standard electrode potentials, calculate the standard free-energy change at [latex]25^{\circ} \mathrm{C}[/latex] for the reaction

[latex]\mathrm{Zn}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{Ag}(s)[/latex]

PROBLEM STRATEGY

Substitute into [latex]\Delta G^{\circ}=-n F E_{ ext {cell }}^{\circ}[/latex]. Use a table of standard electrode potentials to obtain [latex]E_{ ext {cell }}^{\circ}[/latex]. The cell reaction equals the sum of the half-reactions after they have been multiplied by factors so that the electrons cancel in the summation. Note that [latex]n[/latex] is the number of moles of electrons involved in each half-reaction.

Accepted Answer

The half-reactions, corresponding half-cell potentials, and their sums are displayed below.

[latex] \begin{array}{rlr} \mathrm{Zn}(s) & \longrightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{e}^{-} & -E^{\circ}=0.76 \mathrm{~V} \ 2 \mathrm{Ag}^{+}(a q)+2 \mathrm{e}^{-} & \longrightarrow 2 \mathrm{Ag}(s) & E^{\circ}=0.80 \mathrm{~V} \ \hline \mathrm{Zn}(s) +2 \mathrm{Ag}^{+}(a q) & \longrightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{Ag}(s) & E_{ ext {cell }}^{\circ}=1.56 \mathrm{~V} \end{array} [/latex]

Note that each half-reaction involves the transfer of two moles of electrons; hence, [latex]n=2[/latex]. Also, [latex]E_{ ext {cell }}^{\circ}=1.56 \mathrm{~V}[/latex], and the faraday constant, [latex]F[/latex], is [latex]9.65 imes 10^{4} \mathrm{C} / \mathrm{mol} \mathrm{e}^{-}[/latex]. Therefore,

[latex]\Delta G^{\circ}=-n F E_{ ext {cell }}^{\circ}=-2 \mathrm{~mol} \mathrm{e}^{-} imes 96,500 \mathrm{C} / \mathrm{mol} \mathrm{e}^{-} imes 1.56 \mathrm{~V}=-3.01 imes 10^{5} \mathrm{~J}[/latex]

Recall that (coulombs [latex]) imes([/latex] volts [latex])=[/latex] joules. Thus, the standard free-energy change is [latex]\mathbf{- 3 0 1} \mathbf{k J}[/latex].

Question 20.9: Calculating the Free-Energy Change from Electrode Potentials...

Related Answered Questions

Calculating the Cell emf from Free-Energy Change Suppose the reaction of zinc metal and chlorine gas is utilized in a voltaic cell in which zinc ions and chloride ions are formed in aqueous solution. Zn(s) + Cl2(g) → Zn^2+(aq) + 2Cl^-(aq) Calculate the standard emf for this cell at 25°C from ...

Verified Answer:

Verified Answer:

Verified Answer:

Predicting the Half-Reactions in an Aqueous Electrolysis What do you expect to be the half-reactions in the electrolysis of aqueous copper(II) sulfate? ...

Verified Answer:

Verified Answer:

Verified Answer:

Calculating the emf from Standard Potentials Calculate the standard emf of the following voltaic cell at 25°C using standard electrode potentials. Al(s)|Al^3+(aq)||Fe^2+(aq)|Fe(s) What is the cell reaction? PROBLEM STRATEGY From a table of electrode potentials, write the two reduction ...

Verified Answer:

Verified Answer:

Determining the Relative Strengths of Oxidizing and Reducing Agents a. Order the following oxidizing agents by increasing strength under standard-state conditions: Cl2(g), H2O2(aq), Fe^3+(aq). b. Order the following reducing agents by increasing strength under standard-state conditions: H2(g), ...

Verified Answer:

Verified Answer: