Chapter 17, Problem 17.97QA

Interpretation Introduction

To find:

a. Assign oxidation numbers and find out how many electrons are involved in process.

b. Calculate ${∆H}^o$ of the reaction given.

c. Calculate $E^o$ for the reaction.

d. Determine the effect of $∆S$.

e. Assign oxidation numbers and the electrons involved in the process.

Answer to Problem 17.97QA

Solution:

a.  ${In K}_{2}Mn{F}_6;K= +1, Mn= +4, F= -1:in Sb{F}_5;Sb= +5, F= -1:in KSb{F}_6;K= +1, Sb= +5, F= -1:in Mn{F}_3;Mn= +3, F= -1;in F_2;F=0,$ and the number of electrons involved in the process is one.

b. ${∆H}^o of the reaction is-656 kJ/mol$.

c. $E^{o}of the reaction= 6.80 V$

d. The value of $E^o$ of the cell will be too low if $∆S$ is taken into consideration.

e. $In KH{F}_2;K= +1, H= +1, F= -1:in KF;K= +1, F= -1:in H_2;H=0:in F_2;F=0,$ and the number of electrons involved in the process is two.

Explanation of Solution

1) Concept:

We are given a redox reaction and asked to find the oxidation numbers for each elements in the reaction. Oxidation number of an element in the given compound is calculated using oxidation number rules in Table 8.5. In general, the O.N. of pure element is 0, the O.N of a monoatomic ion is the charge on that ion, and the O.N. values of $O$ and $H$ in most common compounds are $0.2$ and $1$ respectively.

The enthalpy of the reaction is the difference between the total enthalpy of formation of products and the total enthalpy of formation of reactants.

Using the equation below and using the standard enthalpy formation values of reactants and products in a given reaction, $∆H_{rxn}^0$ can be calculated.

$∆H_{rxn}^o= \sum n_{products}∆H_{f , products}^o - \sum n_{reactants}∆H_{f, reactants}^o$

Electric work is the product of the quantity of charge flowing and the cell potential of the cell. It is given that $∆G\approx ∆H$, so we can use the following relation to find $E^0$ of the reaction:

$∆G_{cell}^0= -nF{E}_{cell}^o= ∆H_{rxn}^0$

We know that the Gibbs free energy can be calculated by using the formula

$∆G= ∆H-T∆S$

Now, we are considering the value of entropy of the reaction, and it is given that it is greater than zero. This implies that the entropy has a positive value, and when put in the equation, the $T∆S$ value will become more negative. The value of enthalpy for this reaction is also negative, which will increase the total value of the Gibbs free energy.

We also know that the Gibbs free energy is directly proportional to the cell potential of the cell. Thus, if the value of Gibbs free energy is increased, the cell potential will also increase significantly.

2) Formula:

i) $∆H^o= \sum n*∆H_f^o of products - \sum m*∆H_f^o of reactants$

ii) $∆G_{cell}^0= -nF{E}_{cell}^o$

iii) $∆G= ∆H-T∆S$

3) Given:

i) $∆H_f^o of K_{2}Mn{F}_6= -2435 kJ/mol$

ii) $∆H_f^o of Sb{F}_5= -1324 kJ/mol$

iii) $∆H_f^o of KSb{F}_6= -2080 kJ/mol$

iv) $∆H_f^o of Mn{F}_3=- 1579 kJ/mol$

v) Faraday’s Constant $F=96500 C$

4) Calculations:

a. Assigning oxidation numbers for each of the element in the given equation:

Given balanced redox reaction is

$K_{2}Mn{F}_6 \left(s\right)c+2 Sb{F}_5 \left(l\right) \to 2 KSb{F}_6 \left(s\right) + Mn{F}_3 \left(s\right)+\frac{1}{2} F_2 \left(g\right)$

In $K_{2}Mn{F}_6$, let us consider the oxidation number of $Mn$ as x. The oxidation number of potassium ion $K^{+}$ is $+1,$ and that of fluorine ion $F^{-}$ is $-1$. Since the compound carries no charge, the overall sum of the oxidation states should be zero (to make it neutral).

$\therefore 2\left(+1\right)+x+6(-1)=0$

$x+\left(+2\right) +(-6)=0$

$\therefore x= +4$

Therefore, oxidation state of $Mn is+4$.

In $Sb{F}_5$, let us consider the oxidation state of $Sb$ as x. The oxidation state of fluorine ion $F^{-}$ is $-1$. Since the compound carries no charge, the overall sum of the oxidation states should be zero (to make it neutral).

$\therefore x+5\left(-1\right)=0$

$x+\left(-5\right)=0$

$\therefore x= +5$

Therefore, the oxidation state of $Sb is+5$.

In $Mn{F}_3$, since the fluorine ions have oxidation state of $-1$, to make the compound neutral, $Mn$ will have $+3$ oxidation state.

In $KSb{F}_6$, one potassium ion having oxidation state of $+1$ and one fluorine ion of oxidation state $-1$ have been added to $Sb{F}_5$. Thus there is no change in the oxidation state of the $Sb,$ and it remains $+5$.

Fluorine is present in elemental form in $F_2;$ hence the oxidation state is zero.

Thus we can see that $Mn$ from $K_{2}Mn{F}_6$ has an oxidation state change from $+4 \to +3$ by accepting one electron (reduction), and it forms $Mn{F}_3$. Also $F$ in $K_{2}Mn{F}_6$ has an oxidation state change from $-1 \to 0$ by donating one electron (oxidation), and it forms $\frac{1}{2} F_2$. Thus the total number of electrons involved in the transfer is one.

b. The enthalpy of a reaction can be calculated by the following formula

$∆H_{rxn}^o= \sum n_{products}∆H_{f , products}^o - \sum n_{reactants}∆H_{f, reactants}^o$

We have already been given the $∆H_f^o$ of the reactants and products, so we can easily calculate the enthalpy of the reaction using the balanced equation. Since $F_2$ is in its elemental form, its  enthalpy of formation is zero.

The given chemical equation is

$K_{2}Mn{F}_6 \left(s\right)+2 Sb{F}_5 \left(l\right) \to 2 KSb{F}_6 \left(s\right) + Mn{F}_3 \left(s\right)+\frac{1}{2} F_2 \left(g\right)$

We will put the given values of enthalpy of formation in the formula, which is written as

$∆H_{rxn}^o= \sum n_{products}∆H_{f , products}^o - \sum n_{reactants}∆H_{f, reactants}^o$

$∆H^o=\left[ 2\times \left(-\frac{2080kJ}{mol}\right)+\left(1\times -1579\frac{kJ}{mol}\right)\right]-[ \left( 1\times -2435\frac{kJ}{mol}\right)+(2\times \left(-1324\frac{kJ}{mol}\right)]$

$∆H^o=\left[ \left(- 4160\frac{kJ}{mol}\right)+\left( -1579\frac{kJ}{mol}\right)\right]-[ \left( -2435\frac{kJ}{mol}\right)-\left( -2648\frac{kJ}{mol}\right)]$

$∆H^o=\left(-5739\frac{kJ}{mol}\right)-\left( -5083\frac{kJ}{mol}\right)$

$\therefore ∆H^o= -656 kJ/mol$

Thus, the enthalpy of the reaction is $- 656 kJ/mol$.

c. Estimating $E^0$ for the reaction:

Gibbs free energy is calculated by two formulas; one involving enthalpy and entropy of a reaction, and the other one involving cell potential. The formulae for both are shown below:

$∆G^0= ∆H_{rxn}^0-T∆S^0$ and $∆G^0= -nF{E}_{cell}^o$

Since for this reaction, it is assumed that $∆S$ is realtively small and the $∆G \approx ∆H$, the electric energy can now be written as

$∆H_{rxn}^0= -nF{E}_{cell}^o$

We have calculated the enthalpy of the reaction, and we also know the number of electrons being transferred. Therefore, we can find out the cell potential.

By the formula above, $E_{cell}^o$ can be written as

$E_{cell}^o= \frac{∆H}{-nF}$

$E_{cell}^o= \frac{- 656 kJ/mol}{- 1*96500}$

$E_{cell}^0= \frac{656*1000 J/mol}{96500 C}$

$\therefore E_{cell}^o=6.80 V$

Thus, the cell potential of the given reaction is $6.80 V$.

d. We know that the Gibbs free energy can be calculated by using the formula

$∆G= ∆H-T∆S$

Now, we are considering the value of entropy of the reaction, and it is given that it is greater than zero. This implies that the entropy has a positive value, and when put in the equation, the product $T∆S$ will become more negative. The value of enthalpy for this reaction is also negative, which will increase the total value of the Gibbs free energy.

We also know that the Gibbs free energy is directly proportional to the cell potential of the cell. Thus if the value of Gibbs free energy is increased, the cell potential will also increase significantly.

Therefore, the value of cell potential $\left(E_{cell}^0\right)$ that we have calculated for the reaction without considering entropy will be too low when compared to the cell potential calculated when entropy is taken into consideration.

e. Assigning O.N. to each element for the given reaction:

The given reaction is

$2 KH{F}_2 \left(l\right) \to 2 KF \left(l\right)+ H_2 \left(g\right)+ F_2 \left(g\right)$

In $KH{F}_2$, since the fluorine ions have oxidation state of $-1$ and the poattasium ion has oxidation state of $+1$, to make the compound neutral, $H$ will have $+1$ oxidation state.

$H_2$ and $F_2$ are in their elemental form; hence, the oxidation states of $H$ and $F$ are zero.

In $KF$, potassium has $+1$ oxidation satte and fluorine has $-1$ oxidation state, which makes the compound is neutral.

Now, $H in KH{F}_2$ takes one electron to form $\frac{1}{2}{H}_2$, and $F$ gives one electron to form $\frac{1}{2}{F}_2$. To balance the equation, it is multiplied by two on both sides, and hence we get two electron transfer.

Conclusion:

If we can assign the oxidation numbers correctly, we can easily find out which species is undergoing reduction or oxidation and by how many electrons, without the need of half-reactions of the redox reaction.