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

Question

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.

[latex]\mathrm{Zn}(s)+\mathrm{Cl}_{2}(g) \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{Zn}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q)[/latex]

Calculate the standard emf for this cell at [latex]25^{\circ} \mathrm{C}[/latex] from standard free energies of formation (see Appendix C).

PROBLEM STRATEGY

Calculate [latex]\Delta G^{\circ}[/latex] and substitute it and the value of [latex]n[/latex] into the equation [latex]\Delta G^{\circ}=[/latex] [latex]-n F E_{ ext {cell. }}^{\circ}[/latex] Solve for [latex]E_{ ext {cell. }}^{\circ}[/latex]

Accepted Answer

Write the equation with [latex]\Delta G_{f}^{\circ}[/latex] 's beneath.

[latex] \begin{array}{ccccc} \mathrm{Zn}(s)& +\mathrm{Cl}_{2}(g) & \longrightarrow \mathrm{Zn}^{2+}(a q)+ & 2 \mathrm{Cl}^{-}(a q) \ \Delta G_{f}^{\circ}: 0 & 0 & -147.0 & 2 imes(-131.3) \mathrm{kJ} \end{array} [/latex]

Hence,

[latex] \begin{aligned} \Delta G^{\circ} & \left.=\sum n \Delta G_{f}^{\circ} ext { (products } ight)-\sum m \Delta G_{f}^{\circ}( ext { reactants }) \ & =[-147.0+2 imes(-131.3)] \mathrm{kJ} \ & =-410 \mathrm{~kJ}=-4.10 imes 10^{5} \mathrm{~J} \end{aligned} [/latex]

You obtain [latex]n[/latex] by splitting the reaction into half-reactions.

[latex] \begin{aligned} & \mathrm{Zn}(s) \longrightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{e}^{-} \ & \mathrm{Cl}_{2}(g)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cl}^{-}(a q) \end{aligned} [/latex]

Each half-reaction involves the transfer of two moles of electrons, so [latex]n=2[/latex]. Now you substitute into

[latex] \begin{aligned} \Delta G^{\circ} & =-n F E_{ ext {cell }}^{\circ} \ -4.10 imes 10^{5} \mathrm{~J} & =-2 \mathrm{~mol} \mathrm{e}^{-} imes 96,500 \mathrm{C} / \mathrm{mol} \mathrm{e}^{-} imes 10^{4} \mathrm{C} imes E_{ ext {cell }}^{\circ} \end{aligned} [/latex]

Solving for [latex]E_{ ext {cell }}^{\circ}[/latex], you get [latex]\boldsymbol{E}_{ ext {cell }}^{\circ}=\mathbf{2 . 1 2} \mathrm{V}[/latex].

Question 20.10: Calculating the Cell emf from Free-Energy Change Suppose the...

Related Answered Questions

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:

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: