Chapter 17, Problem 17.41QA

Interpretation Introduction

To find:

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

Answer to Problem 17.41QA

Solution:

a)  $E_{cell}^0= -0.478 V$; ${∆G}^0=92.2 kJ$

b)  $E_{cell}^0=0.505 V$;  ${∆G}^0= -97.5kJ$

Explanation of Solution

1) Concept:

To calculate the value of $∆G^0$ and $E_{cell}^0$ for the given two 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. We will calculate the $E_{cell}^0$ of the reaction by using $E_{Ox}^0$ and $E_{red}^0$ values.

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

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

4) Calculations:

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

${Sn}_{\left(aq\right)}^{2+}+\mathit{ }{2e}^{-} \to {Sn}_{\left(s\right)}$ $E_{cathode}^0= -0.136 V$

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

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

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

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

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

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

Calculating $E_{cell}^0$:

${E_{cell}^0= E}_{cathode}^0-E_{anode}^0= -0.136 V-0.342 V= -0.478V$

Calculating ${∆G}^0$:

${∆G}^0= -nF{∆E}_{cell}$         

${{∆\mathrm{G}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}}^0=-2 mol \times \frac{{9.65 \times 10}^4 C}{mol }\times (-0.478 V)$

${{∆\mathrm{G}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}}^0= 9.2254\times {10}^4 C.V=9.2254\times {10}^{4}J= 92.2 kJ$

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

${Ni}_{\left(aq\right)}^{2+}+\mathit{ }{2e}^{-} \to {Ni}_{\left(s\right)}$ $E_{cathode}^0= -0.257 V$

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

${Zn}_{\left(aq\right)}^{2+}+\mathit{ }{2e}^{-} \to {Zn}_{\left(s\right)}$ $E_{anode}^0= -0.762 V$

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

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

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

${Zn}_{\left(s\right)}+{Ni}_{\left(aq\right)}^{2+}\to {Zn}_{\left(aq\right)}^{2+}+ {Ni}_{\left(s\right)}$

Calculating $E_{cell}^0:$

${E_{cell}^0= E}_{cathode}^0-E_{anode}^0= -0.257 V-(-0.762) V= 0.505V$

Calculating ${∆G}^0$:

${∆G}^0= -nF{∆E}_{cell}$         

 ${{∆\mathrm{G}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}}^0=-2 mol \times \frac{{9.65 \times 10}^{4}C}{mol }\times 0.505V$

${{∆\mathrm{G}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}}^0= -9.7465\times {10}^4 C.V=-9.7465\times {10}^4 J= -97.5 kJ$

Conclusion:

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