Question

For each of the following half-cells, compare electrode potentials calculated from (1) concentration and (2) activity data.

(a)$Sn{\left(Cl{O}_4\right)}_2(3.00\times {10}^{-5}M),Sn{\left(Cl{O}_4\right)}_4(6.00\times {10}^{-5}M)|Pt$

(b)$Sn{\left(Cl{O}_4\right)}_2(3.00\times {10}^{-5}M),Sn{\left(Cl{O}_4\right)}_4(6.00\times {10}^{-5}M),NaCl{O}_4(0.0800M)|Pt$

Short answer

The electrode potential calculated from concentration is $0.163V$and that calculated from activity data is $0.140V$.

Step-by-step solution

Concept Introduction

a) The cell given in the question is $Sn{\left(Cl{O}_4\right)}_2(3.00\times {10}^{-5}M),Sn{\left(Cl{O}_4\right)}_4(6.00\times {10}^{-5}M)|Pt$

The half cell reaction and its standard reduction potential

is as follows localid="1645643599661" $S{n}^{4+}+2e^{-}\to S{n}^{2+},E^0=0.154V$

Calculation of electrode potential

The potential of the $E_{S{n}^{2+}}$is as follows:

$E_{S{n}^{2+}}={E^0}_{S{n}^{2+}}-\frac{0.0592}{n}\log \frac{\left[S{n}^{2+}\right]}{\left[S{n}^{4+}\right]}$

Substituting the given values for concentration and the value of ${E^0}_{S{n}^{2+}}$to determine the value of $E_{S{n}^{2+}}$, we get

localid="1651489355744" $$\begin{gathered} E_{S{n}^{2+}}={E^0}_{S{n}^{2+}}-\frac{0.0592}{n}\log \frac{\left[S{n}^{2+}\right]}{\left[S{n}^{4+}\right]} \\ =0.154-\frac{0.0592}{2}\log \frac{3.00\times {10}^{-5}}{6.00\times {10}^{-5}} \\ =0.163V \end{gathered}$$

Calculate of electrode potential from activity data

For calculating the activity of Hydrogen ions, we first calculate the ionic strength $\left(\mu \right)$of the solution using the following formula:$\mu =\frac{1}{2}\left[c_1{Z_1}^2+c_2{Z_2}^2+c_3{Z_3}^2+...\right]$

where localid="1645638883274" $c_1,c_2,c_3...$are molar concentrations of the ions in the solution and $Z_1,Z_2,Z_3,...$are their respective charges. Now, Substituting the molar concentrations of localid="1645639232846" $S{n}^{2+},S{n}^{4+}$and $Cl{O_4}^{-}$and their charges in the above equation, we will get $$\begin{gathered} \mu =\frac{1}{2}\left[\left(3.00\times {10}^{-5}\right){\left(2\right)}^2+(6.00\times {10}^{-5}){\left(4\right)}^2+(6.00\times {10}^{-5}){\left(1\right)}^2+(24\times {10}^{-5}){\left(1\right)}^2\right] \\ =6.9\times {10}^{-4} \end{gathered}$$

Now we shall calculate the activity coefficient of $S{n}^{2+}$using the following equation and substituting the value of ionic strength where ${\alpha }_{S{n}^{2+}}$is the effective diameter of the hydrated ion in nanometers

localid="1651489687862" $$\begin{gathered} -\log {\gamma }_{S{n}^{2+}}=\frac{0.509\times {Z^2}_{S{n}^{2+}}\sqrt{\mu }}{1+(3.28\times {\alpha }_{S{n}^{2+}})\sqrt{\mu }} \\ =\frac{0.509\times {\left(2\right)}^2\sqrt{6.9\times {10}^{-4}}}{1+(3.28\times 0.6)\sqrt{6.9\times {10}^{-4}}} \\ =0.05085 \end{gathered}$$

Hence,localid="1651489733683" ${\gamma }_{S{n}^{2+}}=0.89$

Calculating the activity coefficient of Sn4+

$$\begin{gathered} -\log {\gamma }_{S{n}^{4+}}=\frac{0.509\times {Z^2}_{S{n}^{4+}}\sqrt{\mu }}{1+(3.28\times {\alpha }_{S{n}^{4+}})\sqrt{\mu }} \\ =\frac{0.509\times {\left(4\right)}^2\sqrt{6.9\times {10}^{-4}}}{1+(3.28\times 1.1)\sqrt{6.9\times {10}^{-4}}} \\ =0.195 \end{gathered}$$

and hence ${\gamma }_{S{n}^{4+}}=0.638$

Substituting the values of activity coefficients and localid="1645640909119" $S{n}^{2+}$and $S{n}^{4+}$ion concentrations in the following formula to determine the activities of $S{n}^{2+}$and $S{n}^{4+}$ions, we get

localid="1645641929651" $$\begin{gathered} a_{S{n}^{2+}}={\gamma }_{S{n}^{2+}}{c}_{S{n}^{2+}} \\ =0.890\times (3.00\times {10}^{-5}) \\ =2.67\times {10}^{-5} \\ {a}_{S{n}^{4+}}={\gamma }_{S{n}^{4+}}{c}_{S{n}^{4+}} \\ =0.638\times (6.00\times {10}^{-5}) \\ =3.83\times {10}^{-5} \end{gathered}$$

To Determine ESn2+

Substituting the value of activity of $S{n}^{2+}$and $S{n}^{4+}$in the following formula to determine the $E_{S{n}^{2+}}$,

localid="1645641953946" $$\begin{gathered} E_{S{n}^{2+}}={E^0}_{S{n}^{2+}}-\frac{0.0592}{n}\log \left(\frac{a_{S{n}^{2+}}}{a_{S{n}^{4+}}}\right) \\ =0.154-\frac{0.0592}{2}\log \frac{2.67\times {10}^{-5}}{3.83\times {10}^{-5}} \\ =0.159V \end{gathered}$$

The electrode potential calculated from concentration is $0.163V$and the electrode potential calculated from activity data is$0.159V$.

Concept introduction

b) The given cell is as follows

$Sn{\left(Cl{O}_4\right)}_2(3.00\times {10}^{-5}M),Sn{\left(Cl{O}_4\right)}_4(6.00\times {10}^{-5}M),NaCl{O}_4(0.0800M)/Pt$

The half cell reaction and its standard reduction potential is as follows:

$S{n}^{4+}+2e^{-}\to S{n}^{2+}{E}^0=0.154V$

Calculation of electrode potential from concentration

The potential of the $E_{S{n}^{2+}}$is as follows:

$E_{S{n}^{2+}}={E^0}_{S{n}^{2+}}-\frac{0.0592}{n}\log \frac{\left[S{n}^{2+}\right]}{\left[S{n}^{4+}\right]}$

In the above equation, n is the number of moles of electrons involved in the reaction. By substituting the given values for concentration and the value of ${E^0}_{S{n}^{2+}}$to determine the value of $E_{S{n}^{2+}}$.

localid="1651489893615" $$\begin{gathered} E_{S{n}^{2+}}={E^0}_{S{n}^{2+}}-\frac{0.0592}{n}\log \frac{\left[S{n}^{2+}\right]}{\left[S{n}^{4+}\right]} \\ =0.154-\frac{0.0592}{2}\log \frac{3.00\times {10}^{-5}}{6.00\times {10}^{-5}} \\ =0.163V \end{gathered}$$

Calculation of electrode potential from activity data.

To calculate the activity coefficient of hydrogen ions, first, calculate the ionic strength $\left(\mu \right)$of the solution using the following formula:

$\mu =\frac{1}{2}\left[c_1{Z_1}^2+c_2{Z_2}^2+c_3{Z_3}^2+...\right]$

Where, $c_1,c_2,c_{3,...}$are molar concentrations of the ions in the solution and $Z_1+Z_2+Z_3+...$are their respective charges. The concentrations of $S{n}^{2+},S{n}^{4+}$are very less as compared to the concentration of $Cl{O_4}^{-}$.

Hence, the ionic strength, $\mu$of the solution is due to the chlorate ion and is $0.0800$.

Calculation of the activity coefficients ofSn2+ and Sn4+

$$\begin{gathered} -\log {\gamma }_{S{n}^{2+}}=\frac{0.509\times {Z^2}_{S{n}^{2+}}\sqrt{\mu }}{1+(3.28\times {\alpha }_{S{n}^{2+}})\sqrt{\mu }} \\ =\frac{0.509\times {\left(2\right)}^2\sqrt{0.0800}}{1+(3.28\times 0.6)\sqrt{0.0800}} \\ =0.370 \\ andso,{\gamma }_{S{n}^{2+}}=0.427 \\ Similarly, \\ -\log {\gamma }_{S{n}^{4+}}=\frac{0.509\times {Z^2}_{S{n}^{4+}}\sqrt{\mu }}{1+(3.28\times {\alpha }_{S{n}^{4+}})\sqrt{\mu }} \\ =\frac{0.509\times {\left(4\right)}^2\sqrt{0.0800}}{1+(3.28\times 1.1)\sqrt{0.0800}} \\ =1.140 \\ andso,{\gamma }_{S{n}^{4+}}=0.072 \end{gathered}$$

Substituting the values of activity coefficients of $S{n}^{2+}$and $S{n}^{4+}$ion concentrations in the

following formula to determine the activities of $S{n}^{2+}$and $S{n}^{4+}$, we get

$$\begin{gathered} a_{S{n}^{2+}}={\gamma }_{S{n}^{2+}}{c}_{S{n}^{2+}} \\ =0.427\times (3.00\times {10}^{-5}) \\ =1.28\times {10}^{-5} \\ {a}_{S{n}^{4+}}={\gamma }_{S{n}^{4+}}{c}_{S{n}^{4+}} \\ =0.072\times (6.00\times {10}^{-5}) \\ =4.35\times {10}^{-6} \end{gathered}$$

Determination of ESn2+

Substituting the value of the activity of $S{n}^{2+}$and $S{n}^{4+}$ions in the following formula, we shall determine $E_{S{n}^{2+}}$.

$$\begin{gathered} E_{S{n}^{2+}}={E^0}_{S{n}^{2+}}-\frac{0.0592}{n}\log \left(\frac{a_{S{n}^{2+}}}{a_{S{n}^{4+}}}\right) \\ =0.154-\frac{0.0592}{2}\log \frac{1.28\times {10}^{-5}}{4.35\times {10}^{-5}} \\ =0.140V \end{gathered}$$

The electrode potential calculated from concentration is $0.163V$and The electrode potential calculated from activity data is$0.140V$.