Chapter 19, Problem 64GQ

(a)

Interpretation Introduction

Interpretation:

The element which is the weakest oxidizing agent has to be determined.

Concept introduction:

Under certain conditions a cell potential is measured it is called as standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$.

Standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$ can be calculated by the following formula.

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

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is positive, the reaction is predicted to be product favoured at equilibrium.

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is negative, the reaction is predicted to be reactant favoured at equilibrium.

Electrochemical series:

It is a decreasing order of the reduction potentials. The most positive ${\text{E}}^{\text{0}}$ are placed in the top in the electrochemical series, it has greater tendency for reduction and lower tendency of oxidation. And the most negative ${\text{E}}^{\text{0}}$ values of elements are placed in the bottom of the series.

Let’s write the reduction potential of each of the following non-metal ions.

$\begin{array}{l}{\text{F}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2F}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 2}\text{.87 V}\\ {\text{Cl}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Cl}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.36 V}\\ {\text{Br}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Br}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.08 V}\\ {\text{I}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2I}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 0}\text{.535 V}\end{array}$

Further,

$\begin{array}{l}{\text{O}}_{\text{2}}{\text{+4H}}^{\text{+}}{\text{+ 4e}}^{\text{-}}\to {\text{2H}}_{\text{2}}{\text{O ; E}}^{\text{0}}\text{= + 1}\text{.229 V}\\ {\text{S +2H}}^{\text{+}}{\text{+ 2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{S ; E}}^{\text{0}}\text{= + 0}\text{.14 V}\\ {\text{Se + 2H}}^{\text{+}}{\text{+2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{Se ; E}}^{\text{0}}\text{ = - 0}\text{.40 V}\end{array}$

Answer to Problem 64GQ

$\text{Se}$- Weakest oxidizing agent.

Explanation of Solution

Let’s write the reduction potential of each of the following non-metal ions.

$\begin{array}{l}{\text{F}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2F}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 2}\text{.87 V}\\ {\text{Cl}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Cl}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.36 V}\\ {\text{Br}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Br}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.08 V}\\ {\text{I}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2I}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 0}\text{.535 V}\end{array}$

Further,

$\begin{array}{l}{\text{O}}_{\text{2}}{\text{+4H}}^{\text{+}}{\text{+ 4e}}^{\text{-}}\to {\text{2H}}_{\text{2}}{\text{O ; E}}^{\text{0}}\text{= + 1}\text{.229 V}\\ {\text{S +2H}}^{\text{+}}{\text{+ 2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{S ; E}}^{\text{0}}\text{= + 0}\text{.14 V}\\ {\text{Se + 2H}}^{\text{+}}{\text{+2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{Se ; E}}^{\text{0}}\text{ = - 0}\text{.40 V}\end{array}$

From the given standard electrode potentials $\text{Se}$ has low value.

Therefore, it is weakest oxidizing agent.

(b)

Interpretation Introduction

Interpretation:

The ion or $\text{HX}$ which is the weakest oxidising agent has to be determined.

Concept introduction:

Under certain conditions a cell potential is measured it is called as standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$.

Standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$ can be calculated by the following formula.

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

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is positive, the reaction is predicted to be product favoured at equilibrium.

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is negative, the reaction is predicted to be reactant favoured at equilibrium.

Electrochemical series:

It is an decreasing order of the reduction potentials. The most positive ${\text{E}}^{\text{0}}$ are placed in the top in the electrochemical series, it has greater tendency for reduction and lower tendency of oxidation. And the most negative ${\text{E}}^{\text{0}}$ values of elements are placed in the bottom of the series.

Let’s write the reduction potential of each of the following non-metal ions.

$\begin{array}{l}{\text{F}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2F}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 2}\text{.87 V}\\ {\text{Cl}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Cl}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.36 V}\\ {\text{Br}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Br}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.08 V}\\ {\text{I}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2I}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 0}\text{.535 V}\end{array}$

Further,

$\begin{array}{l}{\text{O}}_{\text{2}}{\text{+4H}}^{\text{+}}{\text{+ 4e}}^{\text{-}}\to {\text{2H}}_{\text{2}}{\text{O ; E}}^{\text{0}}\text{= + 1}\text{.229 V}\\ {\text{S +2H}}^{\text{+}}{\text{+ 2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{S ; E}}^{\text{0}}\text{= + 0}\text{.14 V}\\ {\text{Se + 2H}}^{\text{+}}{\text{+2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{Se ; E}}^{\text{0}}\text{ = - 0}\text{.40 V}\end{array}$

Answer to Problem 64GQ

${\text{F}}^{\text{-}}$- Weakest reducing agent.

Explanation of Solution

Let’s write the reduction potential of each of the following non-metal ions.

$\begin{array}{l}{\text{F}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2F}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 2}\text{.87 V}\\ {\text{Cl}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Cl}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.36 V}\\ {\text{Br}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Br}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.08 V}\\ {\text{I}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2I}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 0}\text{.535 V}\end{array}$

Further,

$\begin{array}{l}{\text{O}}_{\text{2}}{\text{+4H}}^{\text{+}}{\text{+ 4e}}^{\text{-}}\to {\text{2H}}_{\text{2}}{\text{O ; E}}^{\text{0}}\text{= + 1}\text{.229 V}\\ {\text{S +2H}}^{\text{+}}{\text{+ 2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{S ; E}}^{\text{0}}\text{= + 0}\text{.14 V}\\ {\text{Se + 2H}}^{\text{+}}{\text{+2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{Se ; E}}^{\text{0}}\text{ = - 0}\text{.40 V}\end{array}$

From the given standard electrode potentials ${\text{F}}^{\text{-}}$ has high value.

Therefore, it is a weakest reducing agent.

(c)

Interpretation Introduction

Interpretation:

The element which is capable of oxidizing ${\text{H}}_{\text{2}}{\text{O to O}}_{\text{2}}$ has to be determined.

Concept introduction:

Under certain conditions a cell potential is measured it is called as standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$.

Standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$ can be calculated by the following formula.

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

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is positive, the reaction is predicted to be product favoured at equilibrium.

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is negative, the reaction is predicted to be reactant favoured at equilibrium.

Electrochemical series:

It is an decreasing order of the reduction potentials. The most positive ${\text{E}}^{\text{0}}$ are placed in the top in the electrochemical series, it has greater tendency for reduction and lower tendency of oxidation. And the most negative ${\text{E}}^{\text{0}}$ values of elements are placed in the bottom of the series.

Let’s write the reduction potential of each of the following non-metal ions.

$\begin{array}{l}{\text{F}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2F}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 2}\text{.87 V}\\ {\text{Cl}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Cl}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.36 V}\\ {\text{Br}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Br}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.08 V}\\ {\text{I}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2I}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 0}\text{.535 V}\end{array}$

Further,

$\begin{array}{l}{\text{O}}_{\text{2}}{\text{+4H}}^{\text{+}}{\text{+ 4e}}^{\text{-}}\to {\text{2H}}_{\text{2}}{\text{O ; E}}^{\text{0}}\text{= + 1}\text{.229 V}\\ {\text{S +2H}}^{\text{+}}{\text{+ 2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{S ; E}}^{\text{0}}\text{= + 0}\text{.14 V}\\ {\text{Se + 2H}}^{\text{+}}{\text{+2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{Se ; E}}^{\text{0}}\text{ = - 0}\text{.40 V}\end{array}$

Answer to Problem 64GQ

${\text{F}}_{\text{2}}{\text{ and Cl}}_{\text{2}}$ are capable of ${\text{H}}_{\text{2}}{\text{O to O}}_{\text{2}}$

Explanation of Solution

Let’s write the reduction potential of each of the following non-metal ions.

$\begin{array}{l}{\text{F}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2F}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 2}\text{.87 V}\\ {\text{Cl}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Cl}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.36 V}\\ {\text{Br}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Br}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.08 V}\\ {\text{I}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2I}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 0}\text{.535 V}\end{array}$

Further,

$\begin{array}{l}{\text{O}}_{\text{2}}{\text{+4H}}^{\text{+}}{\text{+ 4e}}^{\text{-}}\to {\text{2H}}_{\text{2}}{\text{O ; E}}^{\text{0}}\text{= + 1}\text{.229 V}\\ {\text{S +2H}}^{\text{+}}{\text{+ 2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{S ; E}}^{\text{0}}\text{= + 0}\text{.14 V}\\ {\text{Se + 2H}}^{\text{+}}{\text{+2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{Se ; E}}^{\text{0}}\text{ = - 0}\text{.40 V}\end{array}$

From the given standard electrode potentials ${\text{F}}_{\text{2}}{\text{ and Cl}}_{\text{2}}$ has high E⁰ value.

Therefore, these are strongest oxidizing agents have capable of able to oxidized to water.

(d)

Interpretation Introduction

Interpretation:

The element which is capable of oxidizing ${\text{H}}_{\text{2}}\text{O to S}$ has to be determined.

Concept introduction:

Under certain conditions a cell potential is measured it is called as standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$.

Standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$ can be calculated by the following formula.

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

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is positive, the reaction is predicted to be product favoured at equilibrium.

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is negative, the reaction is predicted to be reactant favoured at equilibrium.

Electrochemical series:

It is an decreasing order of the reduction potentials. The most positive ${\text{E}}^{\text{0}}$ are placed in the top in the electrochemical series, it has greater tendency for reduction and lower tendency of oxidation. And the most negative ${\text{E}}^{\text{0}}$ values of elements are placed in the bottom of the series.

Let’s write the reduction potential of each of the following non-metal ions.

$\begin{array}{l}{\text{F}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2F}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 2}\text{.87 V}\\ {\text{Cl}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Cl}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.36 V}\\ {\text{Br}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Br}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.08 V}\\ {\text{I}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2I}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 0}\text{.535 V}\end{array}$

Further,

$\begin{array}{l}{\text{O}}_{\text{2}}{\text{+4H}}^{\text{+}}{\text{+ 4e}}^{\text{-}}\to {\text{2H}}_{\text{2}}{\text{O ; E}}^{\text{0}}\text{= + 1}\text{.229 V}\\ {\text{S +2H}}^{\text{+}}{\text{+ 2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{S ; E}}^{\text{0}}\text{= + 0}\text{.14 V}\\ {\text{Se + 2H}}^{\text{+}}{\text{+2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{Se ; E}}^{\text{0}}\text{ = - 0}\text{.40 V}\end{array}$

Answer to Problem 64GQ

${\text{F}}_{\text{2}}{\text{ ,Cl}}_{\text{2}}{\text{,Br}}_{\text{2}}{\text{, I}}_{\text{2}}{\text{ and O}}_{\text{2}}$ are able to oxidize ${\text{H}}_{\text{2}}\text{S to S}$.

Explanation of Solution

Let’s write the reduction potential of each of the following non-metal ions.

$\begin{array}{l}{\text{F}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2F}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 2}\text{.87 V}\\ {\text{Cl}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Cl}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.36 V}\\ {\text{Br}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Br}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.08 V}\\ {\text{I}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2I}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 0}\text{.535 V}\end{array}$

Further,

$\begin{array}{l}{\text{O}}_{\text{2}}{\text{+4H}}^{\text{+}}{\text{+ 4e}}^{\text{-}}\to {\text{2H}}_{\text{2}}{\text{O ; E}}^{\text{0}}\text{= + 1}\text{.229 V}\\ {\text{S +2H}}^{\text{+}}{\text{+ 2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{S ; E}}^{\text{0}}\text{= + 0}\text{.14 V}\\ {\text{Se + 2H}}^{\text{+}}{\text{+2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{Se ; E}}^{\text{0}}\text{ = - 0}\text{.40 V}\end{array}$

${\text{F}}_{\text{2}}{\text{ ,Cl}}_{\text{2}}{\text{,Br}}_{\text{2}}{\text{, I}}_{\text{2}}{\text{ and O}}_{\text{2}}$ are able to oxidize ${\text{H}}_{\text{2}}\text{S to S}$.

Because the reduction potential values of ${\text{F}}_{\text{2}}{\text{ ,Cl}}_{\text{2}}{\text{,Br}}_{\text{2}}{\text{, I}}_{\text{2}}{\text{ and O}}_{\text{2}}$ are higher than the ${\text{H}}_{\text{2}}\text{S }$.

(e)

Interpretation Introduction

Interpretation:

It has to be identified whether ${\text{O}}_{\text{2}}$ is capable of oxidizing ${\text{I}}^{\text{-}}{\text{ to I}}_{\text{2}}$ in acid solution.

Concept introduction:

Under certain conditions a cell potential is measured it is called as standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$.

Standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$ can be calculated by the following formula.

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

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is positive, the reaction is predicted to be product favoured at equilibrium.

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is negative, the reaction is predicted to be reactant favoured at equilibrium.

Electrochemical series:

It is an decreasing order of the reduction potentials. The most positive ${\text{E}}^{\text{0}}$ are placed in the top in the electrochemical series, it has greater tendency for reduction and lower tendency of oxidation. And the most negative ${\text{E}}^{\text{0}}$ values of elements are placed in the bottom of the series.

Let’s write the reduction potential of each of the following non-metal ions.

$\begin{array}{l}{\text{F}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2F}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 2}\text{.87 V}\\ {\text{Cl}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Cl}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.36 V}\\ {\text{Br}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Br}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.08 V}\\ {\text{I}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2I}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 0}\text{.535 V}\end{array}$

Further,

$\begin{array}{l}{\text{O}}_{\text{2}}{\text{+4H}}^{\text{+}}{\text{+ 4e}}^{\text{-}}\to {\text{2H}}_{\text{2}}{\text{O ; E}}^{\text{0}}\text{= + 1}\text{.229 V}\\ {\text{S +2H}}^{\text{+}}{\text{+ 2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{S ; E}}^{\text{0}}\text{= + 0}\text{.14 V}\\ {\text{Se + 2H}}^{\text{+}}{\text{+2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{Se ; E}}^{\text{0}}\text{ = - 0}\text{.40 V}\end{array}$

Answer to Problem 64GQ

${\text{O}}_{\text{2}}$ has capable to oxidize ${\text{I}}^{\text{-}}{\text{ to I}}_{\text{2}}$.

Explanation of Solution

Let’s write the reduction potential of each of the following non-metal ions.

$\begin{array}{l}{\text{F}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2F}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 2}\text{.87 V}\\ {\text{Cl}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Cl}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.36 V}\\ {\text{Br}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Br}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.08 V}\\ {\text{I}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2I}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 0}\text{.535 V}\end{array}$

Further,

$\begin{array}{l}{\text{O}}_{\text{2}}{\text{+4H}}^{\text{+}}{\text{+ 4e}}^{\text{-}}\to {\text{2H}}_{\text{2}}{\text{O ; E}}^{\text{0}}\text{= + 1}\text{.229 V}\\ {\text{S +2H}}^{\text{+}}{\text{+ 2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{S ; E}}^{\text{0}}\text{= + 0}\text{.14 V}\\ {\text{Se + 2H}}^{\text{+}}{\text{+2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{Se ; E}}^{\text{0}}\text{ = - 0}\text{.40 V}\end{array}$

Only the stronger oxidizing agent has capable to oxidize ${\text{I}}^{\text{-}}{\text{ to I}}_{\text{2}}$.

From the given non-metals ${\text{O}}_{\text{2}}$ is a strong oxidizing agent. Therefore, ${\text{O}}_{\text{2}}$ has capable to oxidize ${\text{I}}^{\text{-}}{\text{ to I}}_{\text{2}}$.

(f)

Interpretation Introduction

Interpretation:

It has to be identified whether $\text{S}$ is capable of oxidizing ${\text{I}}^{\text{-}}{\text{ to I}}_{\text{2}}$.

Concept introduction:

Under certain conditions a cell potential is measured it is called as standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$.

Standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$ can be calculated by the following formula.

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

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is positive, the reaction is predicted to be product favoured at equilibrium.

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is negative, the reaction is predicted to be reactant favoured at equilibrium.

Electrochemical series:

It is an decreasing order of the reduction potentials. The most positive ${\text{E}}^{\text{0}}$ are placed in the top in the electrochemical series, it has greater tendency for reduction and lower tendency of oxidation. And the most negative ${\text{E}}^{\text{0}}$ values of elements are placed in the bottom of the series.

Let’s write the reduction potential of each of the following non-metal ions.

$\begin{array}{l}{\text{F}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2F}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 2}\text{.87 V}\\ {\text{Cl}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Cl}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.36 V}\\ {\text{Br}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Br}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.08 V}\\ {\text{I}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2I}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 0}\text{.535 V}\end{array}$

Further,

$\begin{array}{l}{\text{O}}_{\text{2}}{\text{+4H}}^{\text{+}}{\text{+ 4e}}^{\text{-}}\to {\text{2H}}_{\text{2}}{\text{O ; E}}^{\text{0}}\text{= + 1}\text{.229 V}\\ {\text{S +2H}}^{\text{+}}{\text{+ 2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{S ; E}}^{\text{0}}\text{= + 0}\text{.14 V}\\ {\text{Se + 2H}}^{\text{+}}{\text{+2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{Se ; E}}^{\text{0}}\text{ = - 0}\text{.40 V}\end{array}$

Answer to Problem 64GQ

$\text{S}$ cannot oxidize  ${\text{I}}^{\text{-}}{\text{ to I}}_{\text{2}}$

Explanation of Solution

Let’s write the reduction potential of each of the following non-metal ions.

$\begin{array}{l}{\text{F}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2F}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 2}\text{.87 V}\\ {\text{Cl}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Cl}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.36 V}\\ {\text{Br}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Br}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.08 V}\\ {\text{I}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2I}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 0}\text{.535 V}\end{array}$

Further,

$\begin{array}{l}{\text{O}}_{\text{2}}{\text{+4H}}^{\text{+}}{\text{+ 4e}}^{\text{-}}\to {\text{2H}}_{\text{2}}{\text{O ; E}}^{\text{0}}\text{= + 1}\text{.229 V}\\ {\text{S +2H}}^{\text{+}}{\text{+ 2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{S ; E}}^{\text{0}}\text{= + 0}\text{.14 V}\\ {\text{Se + 2H}}^{\text{+}}{\text{+2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{Se ; E}}^{\text{0}}\text{ = - 0}\text{.40 V}\end{array}$

Only the stronger oxidizing agent has capable to oxidize ${\text{I}}^{\text{-}}{\text{ to I}}_{\text{2}}$.

$\text{S}$ has low reduction potential. Therefore, $\text{S}$ cannot oxidize  ${\text{I}}^{\text{-}}{\text{ to I}}_{\text{2}}$

(g)

Interpretation Introduction

Interpretation:

It has to be identified whether the reaction ${\text{H}}_{\text{2}}\text{S(aq) + Se(s) }\to {\text{H}}_{\text{2}}\text{Se(aq) + S(s)}$ is product –favoured at equilibrium.

Concept introduction:

Under certain conditions a cell potential is measured it is called as standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$.

Standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$ can be calculated by the following formula.

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

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is positive, the reaction is predicted to be product favoured at equilibrium.

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is negative, the reaction is predicted to be reactant favoured at equilibrium.

Electrochemical series:

It is an decreasing order of the reduction potentials. The most positive ${\text{E}}^{\text{0}}$ are placed in the top in the electrochemical series, it has greater tendency for reduction and lower tendency of oxidation. And the most negative ${\text{E}}^{\text{0}}$ values of elements are placed in the bottom of the series.

Let’s write the reduction potential of each of the following non-metal ions.

$\begin{array}{l}{\text{F}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2F}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 2}\text{.87 V}\\ {\text{Cl}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Cl}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.36 V}\\ {\text{Br}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Br}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.08 V}\\ {\text{I}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2I}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 0}\text{.535 V}\end{array}$

Further,

$\begin{array}{l}{\text{O}}_{\text{2}}{\text{+4H}}^{\text{+}}{\text{+ 4e}}^{\text{-}}\to {\text{2H}}_{\text{2}}{\text{O ; E}}^{\text{0}}\text{= + 1}\text{.229 V}\\ {\text{S +2H}}^{\text{+}}{\text{+ 2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{S ; E}}^{\text{0}}\text{= + 0}\text{.14 V}\\ {\text{Se + 2H}}^{\text{+}}{\text{+2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{Se ; E}}^{\text{0}}\text{ = - 0}\text{.40 V}\end{array}$

Answer to Problem 64GQ

${\text{H}}_{\text{2}}\text{S(aq) + Se(s) }\to {\text{H}}_{\text{2}}\text{Se(aq) + S(s)}$ not a product –favoured at equilibrium.

Explanation of Solution

Let’s write the reduction potential of each of the following non-metal ions.

$\begin{array}{l}{\text{F}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2F}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 2}\text{.87 V}\\ {\text{Cl}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Cl}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.36 V}\\ {\text{Br}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Br}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.08 V}\\ {\text{I}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2I}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 0}\text{.535 V}\end{array}$

Further,

$\begin{array}{l}{\text{O}}_{\text{2}}{\text{+4H}}^{\text{+}}{\text{+ 4e}}^{\text{-}}\to {\text{2H}}_{\text{2}}{\text{O ; E}}^{\text{0}}\text{= + 1}\text{.229 V}\\ {\text{S +2H}}^{\text{+}}{\text{+ 2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{S ; E}}^{\text{0}}\text{= + 0}\text{.14 V}\\ {\text{Se + 2H}}^{\text{+}}{\text{+2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{Se ; E}}^{\text{0}}\text{ = - 0}\text{.40 V}\end{array}$

${\text{H}}_{\text{2}}\text{S(aq) + Se(s) }\to {\text{H}}_{\text{2}}\text{Se(aq) + S(s)}$ from the reaction ${\text{H}}_{\text{2}}\text{S}$ being oxidized to S and $\text{Se}$ being reduced to ${\text{H}}_{\text{2}}\text{Se}$.

To oxidize ${\text{H}}_{\text{2}}\text{S}$, the compound must be a stronger oxidizing agent. But $\text{Se}$ has a lower reduction potential. Therefore, ${\text{H}}_{\text{2}}\text{S(aq) + Se(s) }\to {\text{H}}_{\text{2}}\text{Se(aq) + S(s)}$ not a product –favoured at equilibrium.

(h)

Interpretation Introduction

Interpretation:

It has to be identified whether the reaction ${\text{H}}_{\text{2}}{\text{S(aq) + I}}_2\text{(s) }\to {\text{2H}}^{\text{+}}{\text{(aq) + 2I}}^{\text{-}}\text{(aq)}$ is product –favoured at equilibrium.

Concept introduction:

Under certain conditions a cell potential is measured it is called as standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$.

Standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$ can be calculated by the following formula.

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

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is positive, the reaction is predicted to be product favoured at equilibrium.

The ${\text{E}}_{\text{cell}}^{\text{o}}$ value is negative, the reaction is predicted to be reactant favoured at equilibrium.

Electrochemical series:

It is an decreasing order of the reduction potentials. The most positive ${\text{E}}^{\text{0}}$ are placed in the top in the electrochemical series, it has greater tendency for reduction and lower tendency of oxidation. And the most negative ${\text{E}}^{\text{0}}$ values of elements are placed in the bottom of the series.

Let’s write the reduction potential of each of the following non-metal ions.

$\begin{array}{l}{\text{F}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2F}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 2}\text{.87 V}\\ {\text{Cl}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Cl}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.36 V}\\ {\text{Br}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Br}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.08 V}\\ {\text{I}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2I}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 0}\text{.535 V}\end{array}$

Further,

$\begin{array}{l}{\text{O}}_{\text{2}}{\text{+4H}}^{\text{+}}{\text{+ 4e}}^{\text{-}}\to {\text{2H}}_{\text{2}}{\text{O ; E}}^{\text{0}}\text{= + 1}\text{.229 V}\\ {\text{S +2H}}^{\text{+}}{\text{+ 2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{S ; E}}^{\text{0}}\text{= + 0}\text{.14 V}\\ {\text{Se + 2H}}^{\text{+}}{\text{+2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{Se ; E}}^{\text{0}}\text{ = - 0}\text{.40 V}\end{array}$

Answer to Problem 64GQ

${\text{H}}_{\text{2}}{\text{S(aq) + I}}_2\text{(s) }\to {\text{2H}}^{\text{+}}{\text{(aq) + 2I}}^{\text{-}}\text{(aq)}$ product favoured at equilibrium.

Explanation of Solution

Let’s write the reduction potential of each of the following non-metal ions.

$\begin{array}{l}{\text{F}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2F}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 2}\text{.87 V}\\ {\text{Cl}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Cl}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.36 V}\\ {\text{Br}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2Br}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 1}\text{.08 V}\\ {\text{I}}_{\text{2}}{\text{+2e}}^{\text{-}}\to {\text{2I}}^{\text{-}}{\text{ ; E}}^{\text{0}}\text{= + 0}\text{.535 V}\end{array}$

Further,

$\begin{array}{l}{\text{O}}_{\text{2}}{\text{+4H}}^{\text{+}}{\text{+ 4e}}^{\text{-}}\to {\text{2H}}_{\text{2}}{\text{O ; E}}^{\text{0}}\text{= + 1}\text{.229 V}\\ {\text{S +2H}}^{\text{+}}{\text{+ 2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{S ; E}}^{\text{0}}\text{= + 0}\text{.14 V}\\ {\text{Se + 2H}}^{\text{+}}{\text{+2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{Se ; E}}^{\text{0}}\text{ = - 0}\text{.40 V}\end{array}$

${\text{H}}_{\text{2}}{\text{S(aq) + I}}_2\text{(s) }\to {\text{2H}}^{\text{+}}{\text{(aq) + 2I}}^{\text{-}}\text{(aq)}$ from the reaction ${\text{H}}_{\text{2}}\text{S}$ is being oxidized to $\text{S}$ and ${\text{I}}_{\text{2}}$ is being reduced to ${\text{I}}^{\text{-}}$.

To oxidize ${\text{H}}_{\text{2}}\text{S}$, the compound must be a stronger oxidizing agent. This means that it must be easier to reduce .${\text{I}}_{\text{2}}$ has a higher reduction potential. Therefore, ${\text{H}}_{\text{2}}{\text{S(aq) + I}}_2\text{(s) }\to {\text{2H}}^{\text{+}}{\text{(aq) + 2I}}^{\text{-}}\text{(aq)}$ from the reaction ${\text{H}}_{\text{2}}\text{S}$ is being oxidized to $\text{S}$ and ${\text{I}}_{\text{2}}$ is  being reduced to ${\text{I}}^{\text{-}}$.