Chapter 17, Problem 17.42QA

Interpretation Introduction

To find:

The value of  $E_{cell}^0$ and ${∆G}^0$ for the given two redox reactions by using the standard reduction potential listed in Appendix 6 of the text.

 

Answer to Problem 17.42QA

Solution:

a) $E_{cell}^0= 0.789 V$; ${∆G}^0=-152 kJ$

b) $E_{cell}^0=-0.03V$;  ${∆G}^0= 2.90 kJ$

 

Explanation of Solution

1) Concept:

To calculate the value of $∆G^0$ and $E_{cell}^0$ for the two given reactions, we can use the appropriate standard potentials listed in Appendix 6.  First, we will calculate the $E_{Ox}^0$ and $E_{red}^0$ from the balanced reaction. Then, we will calculate the $E_{cell}^0$ of the reaction by using $E_{Ox}^0$ and $E_{red}^0$ values.

Form the $E_{cell}^0$ and $∆G^0$ formula we can calculate $∆G^0$.

2) Formula:

i) $E_{cell}^0= E_{cathod}^0-E_{anode}^0$    

ii) ${∆G}^0= -nF{∆E}_{cell}$         

      

3) Given:

i) ${Fe}_{\left(s\right)}+{Cu}_{\left(aq\right)}^{2+}\to {Fe}_{\left(aq\right)}^{2+}+ {Cu}_{\left(s\right)}$    

ii) ${Ag}_{\left(s\right)}+{Fe}_{\left(aq\right)}^{2+}\to {Ag}_{\left(aq\right)}^{+}+ {Fe}_{\left(aq\right)}^{2+}$

4) Calculations:

a) The half reaction at the cathode is based on the reduction of ${Cu}^{2+}$ to $Cu$. The appropriate half-reaction in Table A6.1 is

${Cu}_{\left(aq\right)}^{2+}+\mathit{ }{2e}^{-} \to {Cu}_{\left(s\right)}$ $E_{anode}^0= 0.342 V$

We must reverse the anode’s oxidation half-reaction to find an entry in Table A6.1 in which ${Fe}^{2+}$ is the reactant and Fe is the product:

${Fe}_{\left(aq\right)}^{2+}+\mathit{ }{2e}^{-} \to {Fe}_{\left(s\right)}$ $E_{anode}^0= -0.447 V$

Reversing the ${Fe}^{2+}$half reaction and adding it to the ${Cu}^{2+}$ half reaction, we get

${Cu}_{\left(aq\right)}^{2+}+\mathit{ }{2e}^{-} \to {Cu}_{\left(s\right)}$

${Fe}_{\left(s\right)} \to {Fe}_{\left(aq\right)}^{2+}+\mathit{ }{2e}^{-}$

 

${Fe}_{\left(s\right)}+{Cu}_{\left(aq\right)}^{2+}\to {Fe}_{\left(aq\right)}^{2+}+ {Cu}_{\left(s\right)}$    

Calculating $E_{cell}^0$,

${E_{cell}^0= E}_{cathod}^0-E_{anode}^0= 0.342 V-\left(-0.447 V\right)= 0.789 V$

Calculating ${∆G}^0$,

${∆G}^0= -nF{∆E}_{cell}$         

${{∆\mathrm{G}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}}^0=-2 mol \times \frac{{9.65 \times 10}^{4}C}{mol }\times 0.789 V$

${{∆\mathrm{G}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}}^0= -15.2277\times {10}^{4}C.V=-15.2277\times {10}^{4}J= -152 kJ$

b) The half reaction at the cathode is based on the reduction of ${Fe}^{3+}$ to ${Fe}^{2+}$. The appropriate half-reaction in Table A6.1 is

${Fe}_{\left(aq\right)}^{3+}+\mathit{ }{e}^{-} \to {{Fe}^{2+}}_{\left(s\right)}$ $E_{cathod}^0= 0.770 V$

We must reverse the anode’s oxidation half-reaction to find an entry in Table A6.1 in which ${Ag}^{+}$ is the reactant and Ag is the product:

${Ag}_{\left(aq\right)}^{+}+\mathit{ }{e}^{-} \to {Ag}_{\left(s\right)}$ $E_{anode}^0= 0.800 V$

Reversing the ${Ag}^{+}$half reaction and adding it to  ${Fe}^{3+}$  half reaction, we get

${Fe}_{\left(aq\right)}^{3+}+\mathit{ }{e}^{-} \to {{Fe}^{2+}}_{\left(s\right)} \mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }$

${Ag}_{\left(aq\right)}^{+}+\mathit{ }{e}^{-} \to {Ag}_{\left(s\right)}$

${Ag}_{\left(s\right)}+{Fe}_{\left(aq\right)}^{2+}\to {Ag}_{\left(aq\right)}^{+}+ {Fe}_{\left(aq\right)}^{2+}$

Calculating $E_{cell}^0$,

${E_{cell}^0= E}_{cathod}^0-E_{anode}^0= 0.770 V-0.800 V= -0.03V$

Calculating ${∆G}^0$,

${∆G}^0= -nF{∆E}_{cell}$         

${{∆\mathrm{G}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}}^0=-1 mol \times \frac{{9.65 \times 10}^{4}C}{mol }\times -0.03V$

${{∆\mathrm{G}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}}^0= 0.2895\times {10}^{4}C.V=0.2895\times {10}^{4}J= 2.90 kJ$

Conclusion:

We calculate the $∆G^0$ and $E_{cell}^0$ by using the standard reduction potential values and formula between $∆G^0$ and $E_{cell}^0.$