Chapter 18, Problem 18.81QP

Interpretation Introduction

Interpretation:

The amount of magnesium present at the end of the reaction and molar concentration of ${\text{Ag}}^{+}$ and ${\text{Mg}}^{2+}$ at the equilibrium conditions has to be found.

Concept Introduction:

The standard electrode potential of a cell ${\text{(E}}^{°}{}_{\text{cell}})$ is the difference in electrode potential of the cathode and anode.

${\text{E}}^{\text{°}}{}_{\text{cell}}={\text{E}}^{\text{°}}{}_{\text{cathode}}-{\text{E}}^{\text{°}}{}_{\text{anode}}$

The relation between electrode potential and equilibrium constant: cell potential and equilibrium constant are related by the given equation.

${\text{E}}^{°}{}_{\text{cell}}\text{=}\frac{\text{RT}}{\text{nF}}\text{lnK}$

Where,

${\text{E}}^{°}{}_{\text{cell}}$ is the standard electrode potential

$\text{n}$ is the number of electrons

$\text{K}$ is the formation constant

$\text{R}$ is the universal gas constant $\text{(R=8}{\text{.314JK}}^{\text{-1}}{\text{mol}}^{\text{-1}})$

$\text{T}$ is the temperature

$\text{F}$ is the Faraday constant $\text{(F=9}\text{.64853399}\times {\text{10}}^4{\text{Cmol}}^{\text{-1}})$

Mole is the mass of the substance that is having Avogadro number of particles.  The number of moles of a given substance can be calculated by multiplying its volume with concentration.

Molar concentration or molarity is the ratio of number of moles of solute to the volume of the solution in litre.

Answer to Problem 18.81QP

• The amount of magnesium present at the end of the reaction $=\text{1}\text{.44g}\text{Mg}$
• The molar concentration of ${\text{Ag}}^{\text{+}}$ at equilibrium condition was found to be $=7\times {10}^{-55}\text{M}$
• The molar concentration of ${\text{Mg}}^{2+}$ at equilibrium condition was found to be $=0.0500\text{M}$

Explanation of Solution

The total volume of the ${\text{AgNO}}_3$ solution $=\text{100mL}$

The concentration of ${\text{AgNO}}_3$ $=\text{0}\text{.100M}$

Weight of magnesium taken $=\text{1}\text{.56g}$

$\begin{array}{l}\text{Temperature}\text{= }{\text{25}}^{\text{°}}\text{C}\\ \text{=}\text{25}\text{+}\text{273}\\ \text{=}\text{298K}\end{array}$

Molecular weight of magnesium $=\text{24}\text{.31g}$

To calculate the number of moles of $\text{Mg}$

The number of moles of $\text{Mg}$ can be calculated by dividing the weight of magnesium to its molecular weight.

$\frac{\text{1}\text{.56g}\text{Mg}}{\text{24}\text{.31g}\text{Mg}}\text{=0}\text{.0642}\text{mol}\text{Mg}$

To calculate the number of moles of ${\text{Ag}}^{+}$

The number of moles of ${\text{Ag}}^{+}$ can be calculated by dividing the weight of silver to its molecular weight.

$\text{0}\text{.100}\text{}\text{ mol}{\text{Ag}}^{+}\times \text{0}\text{.1000L=}\text{0}\text{.0100}\text{mol}{\text{Ag}}^{+}$

To calculate the amount of $\text{Mg}$consumed in the given reaction.

The amount of $\text{Mg}$ consumed was found as given below.

$\text{0}\text{.100}\text{}\text{ mol}{\text{Ag}}^{+}\times \frac{\text{1}\text{molMg}}{\text{2}\text{mol}{\text{Ag}}^{\text{+}}}\text{=}\text{0}\text{.00500}\text{mol}\text{Mg}$

To find the amount of $\text{Mg}$remaining in the solution.

The amount of $\text{Mg}$ remaining in the solution is calculated as given below.

$\text{(0}\text{.0642-0}\text{.00500})\text{mol}\text{Mg}\times \frac{\text{24}\text{.31}\text{g Mg}}{1\text{mol}{\text{Ag}}^{\text{+}}}\text{=}\text{1}\text{.44g}\text{Mg}$

To find the concentration of ${\text{Mg}}^{2+}$produced.

The concentration ${\text{Mg}}^{2+}$ produced can be calculated as given below.

$\begin{array}{l}\left[{\text{Mg}}^{2+}\right]=\frac{\text{0}\text{.00500}\text{mol}}{\text{0}\text{.100L}}\\ =0.0500\text{M}\end{array}$

To find ${\text{E}}^{\text{°}}{}_{\text{cell}}$

The standard cell potential can be calculated as given below

$\begin{array}{l}{\text{E}}^{\text{°}}{}_{\text{cell}}={\text{E}}^{\text{°}}{}_{\text{cathode}}-{\text{E}}^{\text{°}}{}_{\text{anode}}\\ =0.80\text{V-(-2}\text{.37V)}\\ =3.17\text{V}\end{array}$

To calculate the equilibrium constant.

The equilibrium constant and the electrode potential is related by the given equation.

${\text{E}}^{°}{}_{\text{cell}}\text{=}\frac{\text{RT}}{\text{nF}}\text{lnK}$

On rearranging we get,

$\begin{array}{l}\text{lnK }=\frac{{\text{nFE}}^{\text{°}}{}_{\text{cell}}}{\text{RT}}\\ =\frac{{\text{(2)(96500JVmol}}^{\text{-1}}\text{)(3}\text{.17V)}}{\text{(8}{\text{.314JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)}}\end{array}$

$\begin{array}{l}\text{K}\text{=}{\text{e}}^{\frac{{\text{(2)(96500JVmol}}^{\text{-1}}\text{)(3}\text{.17V)}}{\text{(8}{\text{.314JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)}}}\\ =1\times {\text{10}}^{107}\end{array}$

To find the equilibrium concentration of ${\text{Mg}}^{2+}$and ${\text{Ag}}^{+}$

Assume that the concentration of ${\text{Mg}}^{2+}$ at equilibrium is $\text{x}$,  the concentration of ${\text{Ag}}^{+}$ will be $\text{2x}$ and the initial concentration of ${\text{Ag}}^{+}$ is zero.

$\begin{array}{l}{\text{2Ag}}^{\text{+}}{}_{\text{(aq)}}{\text{ + Mg}}_{\text{(s)}}\to {\text{ 2Ag}}_{\text{(s)}}{\text{ + Mg}}^{\text{2+}}{}_{\text{(aq)}}\\ \\ \text{Initial(M):0}\text{.0}\text{0}\text{.0500}\\ \text{Change(M):+2x}\text{-x}\end{array}$

$\text{Equilibrium(M)}:\text{2x}\left(\text{0}\text{.0500-x}\right)$

The equilibrium constant,

$\text{K=}\frac{\left[{\text{Mg}}^{\text{2+}}\right]}{{\left[{\text{Ag}}^{\text{2+}}\right]}^{\text{2}}}$

$\text{1}\times {\text{10}}^{107}\text{=}\frac{\left(\text{0}\text{.0500-x}\right)}{{\left(\text{2x}\right)}^{\text{2}}}$

Assume that,                                    $\text{0}\text{.0500-x}\approx 0.0500$

$\text{1}\times {\text{10}}^{107}\approx \frac{\text{0}\text{.0500}}{{\left(\text{2x}\right)}^{\text{2}}}$ We get,

On solving we can find the value of  $\text{x}$

$\begin{array}{l}{\left(\text{2x}\right)}^{\text{2}}=\frac{\text{0}\text{.0500}}{\text{1}\times {\text{10}}^{107}}\\ =0.0500\times {10}^{-107}\end{array}$

$\begin{array}{l}{\left(\text{2x}\right)}^{\text{2}}=5.00\times {10}^{-109}\\ =50.0\times {10}^{-110}\\ 2x=\sqrt{50.0\times {10}^{-110}}\\ =7\times {10}^{-55}\text{M}\end{array}$

Hence,

The concentration of ${\text{Ag}}^{+}$ $=2x$ $=7\times {10}^{-55}\text{M}$

The concentration of ${\text{Mg}}^{2+}$ $=\text{0}\text{.0500-x}=0.0500\text{M}$

Conclusion

The amount of magnesium present at the end of the reaction and molar concentration of ${\text{Ag}}^{+}$ and ${\text{Mg}}^{2+}$ at equilibrium has been found.