Chapter 13, Problem 13.49PAE

Calculate the equilibrium constant for the redox reactions that could occur in the following situations and use that value to explain whether or not any reaction will be observed.

(a) A piece of iron is placed in a 1.0 M solution of NiCl₂(aq).

(b) A copper wire is placed in a 1.0 M solution of Pb(NO₃)₂(aq).

Interpretation Introduction

Interpretation: The equilibrium constant for the redox reaction should be determined along with by using the value of equilibrium constant identifies whether the reaction occurs or not.

A piece of iron is placed in a 1.0 M solution of ${\mathrm{NiCl}}_2$ $\left(aq\right)$.

Concept introduction: The equilibrium constant ${\text{(K}}_{\text{c}}\text{)}$ for a reaction is the ratio of the product of the concentration of the products raised to the power of their stoichiometric coefficients to the product of the concentration of the reactants raised to their power of the stoichiometric coefficients. The redox reaction included with electron transfer process. All chemical reactions included with redox reactions. Both oxidation and reduction occurs in redox reactions.

Answer to Problem 13.49PAE

Solution:

K_c=0.4407 and reaction will occur.

Explanation of Solution

An iron piece is placed in $\text{1}{\text{.0 m NiCl}}_{\text{2}}\left(\text{aq}\right)$

$\text{Fe(s) + NiCl(aq)}\to {\text{FeCl}}_{\text{2}}\text{(aq) + Ni(s)}$

Anode:

$\text{Fe(s)}\to {\text{Fe}}^{\text{2+}}{\text{+2e}}^{-}$

Cathode:

${\text{Ni}}^{2+}{\text{(aq)+2e}}^{-}\to \text{Ni(s)}$

$\begin{array}{l}{\text{E}}^{\text{0}}{}_{{\text{Fe}}^{\text{+2}}\text{/Fe }}\text{= -0}\text{.44 V}\\ {\text{E}}^{\text{0}}{}_{{\text{Ni}}^{\text{+2}}\text{/Ni}}\text{ = -0}\text{.25 V}\end{array}$

Formula used:

${\text{E}}_{\text{cell}}^{\text{0}}\text{=}\frac{\text{0}\text{.059}}{\text{n}}{\text{logK}}_{\text{c}}$

where,

${\text{E}}_{\text{cell}}^{\text{0}}$= reduction potential

n =number of electrons

K_c= equilibrium constant

Calculation: Cathode and anode is identified from above equation

$\begin{array}{l}{\text{E}}^0{}_{\text{cell}}={\text{E}}^0{}_{\text{cathode}}-{\text{E}}^{\text{0}}{}_{\text{anode}}\\ \text{= -0}\text{.25-(-0}\text{.44)}\\ \text{= -0}\text{.25+0}\text{.44}\\ \text{= 0}\text{.19 V}\end{array}$

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{0}}\text{=}\frac{\text{0}\text{.059}}{\text{n}}{\text{logK}}_{\text{c}}\\ 0.19=\frac{\text{0}\text{.059}}{2}\log {\text{K}}_{\text{c}}\\ \mathrm{l}{\text{ogK}}_{\text{c}}=\frac{2\times 0.19}{0.059}=6.4407\\ {\text{K}}_{\text{c}}=\text{antilog}(6.4407)\\ {\text{K}}_{\text{c}}=0.4407\end{array}$

The value of equilibrium constant is positive which implies product is more and reaction will occur.

Thus, reaction will occur.

Interpretation Introduction

Interpretation: The equilibrium constant for the redox reaction should be determined along with by using the value of equilibrium constant identifies whether the reaction occurs or not.

A copper wire is placed in a 1.0 M solution of ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{(aq)}$.

Concept introduction: The equilibrium constant ${\text{(K}}_{\text{c}}\text{)}$ for a reaction is the ratio of the product of the concentration of the products raised to the power of their stoichiometric coefficients to the product of the concentration of the reactants raised to their power of the stoichiometric coefficients. The redox reaction included with electron transfer process. All chemical reactions included with redox reactions. Both oxidation and reduction occurs in redox reactions.

Answer to Problem 13.49PAE

Solution:

K_c=1.175×10-16 and reaction will not occur.

Explanation of Solution

${\text{Cu(s)+Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}\to {\text{Cu}}^{\text{2+}}\text{+Pb(s)}$

Anode:

$\text{Cu(s)}\to {\text{Cu}}^{\text{2+}}{\text{(aq)+2e}}^{-}$

Cathode:

${\text{Pb}}^{2+}{\text{(aq)+2e}}^{-}\to \text{Pb(s)}$

$\begin{array}{l}{\text{E}}^0{}_{{\text{Cu}}^{+2}/\text{Cu}}\text{ = 0}\text{.34 V}\\ {\text{E}}^0{}_{{\text{Pb}}^{\text{+2}}\text{/Pb}}\text{ = -0}\text{.13 V}\end{array}$

Formula used:

${\text{E}}_{\text{cell}}^{\text{0}}\text{=}\frac{\text{0}\text{.059}}{\text{n}}{\text{logK}}_{\text{c}}$

where,

${\text{E}}_{\text{cell}}^{\text{0}}$= reduction potential

n =number of electrons

K_c= equilibrium constant

Calculation: Cathode and anode is identified from the above equation.

$\begin{array}{l}{\text{E}}^0{}_{\text{cell}}={\text{E}}^0{}_{\text{cathode}}-{\text{E}}^{\text{0}}{}_{\text{anode}}\\ \text{= -0}\text{.13 - 0}\text{.34}\\ \text{= -0}\text{.47 V}\end{array}$

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{0}}\text{=}\frac{\text{0}\text{.059}}{\text{n}}{\text{logK}}_{\text{c}}\\ -0.47=\frac{\text{0}\text{.059}}{2}\log {\text{K}}_{\text{c}}\\ \mathrm{l}{\text{ogK}}_{\text{c}}=\frac{2\times -0.47}{0.059}=-15.93\\ {\text{K}}_{\text{c}}\text{=antil}og(-15.93)\\ {\text{K}}_{\text{c}}=1.175\times {10}^{-16}\end{array}$

In this case, the value of equilibrium constant is so small and reactants predominates over products.

Thus, reaction will not occur.