Chapter 17, Problem 17.36QA

Interpretation Introduction

To calculate:

The $∆{\mathrm{G}}^0$ and  ${∆\mathrm{E}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}^0$ of the given reactions

Answer to Problem 17.36QA

Solution:

a) $∆G^0=18.0 kJ; E_{cell}^0=- 0.093 V$

b)  $∆G^0=-611.4 kJ; E_{cell}^0=1.58 V$

Explanation of Solution

1) Concept:

We are asked to calculate the value of $∆G^0$ and $E_{cell}^0$ for the given two reactions using appropriate standard free energies of formation from Appendix 4. Multiplying the $∆{\mathrm{G}}_{\mathrm{f}}^0$ values by the appropriate numbers of moles and subtracting the sum of the reactants values from the sum of the product values given by the equation below will yield the value for $\mathrm{ }∆{\mathrm{G}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0$.

$∆{\mathrm{G}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0=\mathrm{ }\sum {\mathrm{n}}_{\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{t}\mathrm{s}}∆{\mathrm{G}}_{\mathrm{f}\mathrm{ }\left(\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{t}\mathrm{s}\right)}^0-\sum \mathrm{ }{\mathrm{n}}_{\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{s}}∆{\mathrm{G}}_{\mathrm{f}\mathrm{ }\left(\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{s}\right)}^0\mathrm{ }$

Relation between change in standard free energy and standard cell potential is

$∆G^0=- nF{E}_{cell}^0$, where, the value of Faraday constant (F) is  $\frac{{9.65 \times 10}^{4}C}{mol}$ and n is the mole of electrons transferred in the half reactions of the cell.

2) Formula:

i) $∆G^0=\left[\sum n_{product}∆G_{product}^0\right]-\left[\sum n_{reactant}∆G_{reactant}^0\right]$

ii)  $∆G^0=- nF{E}_{cell}^0$     

      

3) Given:

i)   ${FeO}_{ }\left(s\right)+ H_{2 }\left(g\right)\to {Fe}_{ }\left(s\right)+H_2{O}_{ }\left(l\right)$      

ii) ${2Pb}_{ }\left(s\right)+ O_2\left(g\right)+2H_2{SO}_4\left(aq\right)\to {2PbSO}_4\left(s\right)+{2H}_2{O}_{ }\left(l\right)$

    

4) Calculations:

a) ${FeO}_{ }\left(s\right)+ H_{2 }\left(g\right)\to {Fe}_{ }\left(s\right)+H_2{O}_{ }\left(l\right)$      

From the Appendix 4, the values of $∆G_f^0$ are as under:

	$∆{\mathrm{G}}_{\mathrm{f}}^0(\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l})$
$\mathrm{F}\mathrm{e}\mathrm{O}\mathrm{ }\left(\mathrm{s}\right)$	$-255.2$
$H_2\mathrm{ }\left(\mathrm{g}\right)$	$0$
$\mathrm{F}\mathrm{e}\mathrm{ }\left(\mathrm{s}\right)\mathrm{ }$	$0$
$H_{2}O \left(l\right)$	$-237.2$

$∆G^0=\left[\sum n_{product}∆{G_f}_{product}^0\right]-\left[\sum n_{reactant}∆{G_f}_{reactant}^0\right]$

$∆G^0= \left[\left[1 mol Fe \times \left(0 \frac{kJ}{mol}Fe\right)\right]+\left[1 mole H_{2}O \times \left(-237.2 \frac{kJ}{mol}{H}_{2}O\right)\right]\right]-\left[\left[1 mol FeO\times \left(-255.2 \frac{kJ}{mol} FeO\right)\right]+\left[1 mol H_2\times \left(0 \frac{kJ}{mol} H_2\right)\right]\right]$

$∆G^0=-237.2 kJ+255.2 kJ$

$∆G^0= 18.0 kJ$

$∆G^0= 18.0 kJ \times \frac{{10}^{3}J}{1 kJ}= 1.8 \times {10}^4 J$

The number of moles of electrons transferred in the reaction is 2 moles, so n = 2.

$E_{cell}^0$ is calculated as

$∆G^0=-nF{∆E}_{cell}^0$     

$1.8 \times {10}^4 J=-\left( 2 mol \times \frac{{9.65 \times 10}^{4}C}{mol} \times E_{cell}^0\right)$     

${1.8 \times {10}^4 J = -2 \times 9.65 \times 10}^{4}C \times E_{cell}^0$     

$E_{cell}^0=- 0.093 V$     

b)

${2Pb}_{ }\left(s\right)+ O_2\left(g\right)+2H_2{SO}_4\left(aq\right)\to {2PbSO}_4\left(s\right)+{2H}_2{O}_{ }\left(l\right)$

From the Appendix 4, the values of $∆G_f^0$ are

	$∆{\mathrm{G}}_{\mathrm{f}}^0(\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l})$
$\mathrm{P}\mathrm{b}\mathrm{ }\left(\mathrm{s}\right)$	$0$
$O_2\mathrm{ }\left(\mathrm{g}\right)$	$0$
${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\mathrm{ }\left(\mathrm{a}\mathrm{q}\right)\mathrm{ }$	$-744.5$
$PbS{O}_4 \left(s\right)$	$-813.0$
$H_{2}O \left(l\right)$	$-237.2$

$∆G^0=\left[\sum n_{product}∆{G_f}_{product}^0\right]-\left[\sum n_{reactant}∆{G_f}_{reactant}^0\right]$

$∆G^0= \left[\left[2 mol {PbSO}_4 \times \left(-813.0\frac{kJ}{mol} {PbSO}_4\right)\right]+\left[2 mole H_{2}O \times \left(-237.2\frac{kJ}{mol}{H}_{2}O\right)\right]\right]-\left[2 mol H_2{SO}_4\times \left(-744.5 \frac{kJ}{mol} H_2{SO}_4\right)+ 2 mole Pb \times \left(0\frac{kJ}{mol}Pb\right)+1 mol O_2\times \left(0\frac{kJ}{mol}{O}_2\right)\right]$

$∆G^0=-1626.0 kJ-474.4 kJ+1489.0 kJ-0 kJ-0 kJ$

$∆G^0=-611.4 kJ$

$∆G^0= -611.4 kJ\times \frac{{10}^{3}J}{1 kJ}= -6.114\times {10}^5 J$

The number of moles of electrons transferred in the reaction is 4 moles, so n = 4.

$E_{cell}^0$ is calculated as

$∆G^0=-nF{E}_{cell}^0$     

$-6.114\times {10}^5 J=-\left( 4 mol \times \frac{{9.65 \times 10}^{4}C}{mol} \times E_{cell}^0\right){-6.114\times {10}^5 J = - 4 \times 9.65 \times 10}^{4}C \times E_{cell}^0$     

$E_{cell}^0=1.58 V$     

Conclusion:

When the cell reaction is spontaneous, the value of $∆G^0$ is negative and the value of $E_{cell}^0$ is positive. When the cell reaction is non-spontaneous, the value of $∆G^0$ is positive and the value of $E_{cell}^0$ is negative.