Chapter 16, Problem 16.1P

Interpretation Introduction

Interpretation:

Values of $E^{\text{o}}$ and $K$ for below titration reaction have to be determined.

  ${\text{Ce}}^{4+}+{\text{Fe}}^{2+}\to {\text{Ce}}^{3+}+{\text{Fe}}^{3+}$

Concept Introduction:

Expression to calculate value of $K$ is as follows:

  $K={10}^{n{E}^{\text{o}}\text{/0}\text{.05916}}$        (1)

Here,

$n$ is number of electrons.

$E^{\text{o}}$ is standard cell potential.

Answer to Problem 16.1P

Values of $E^{\text{o}}$ and $K$ are $0.933\text{ V}$ and $6\times {10}^{15}$.

Explanation of Solution

Given titration reaction is as follows:

  ${\text{Ce}}^{4+}+{\text{Fe}}^{2+}\to {\text{Ce}}^{3+}+{\text{Fe}}^{3+}$

Reduction half-cell reaction of electrochemical cell is as follows:

  ${\text{Ce}}^{4+}+{\text{e}}^{-}\to {\text{Ce}}^{3+}$

Reduction potential of right half-cell reaction is $1.70\text{ V}$.

Oxidation half-cell reaction of electrochemical cell is as follows:

  ${\text{Fe}}^{2+}\to {\text{Fe}}^{3+}+{\text{e}}^{-}$

Reduction potential of left half-cell reaction is $0.767\text{ V}$.

Value of $E^{\text{o}}$ can be calculated as follows:

  $\begin{array}{c}{E}^{\text{o}}=E_{\text{Red}}^{\text{o}}-E_{\text{Ox}}^{\text{o}}\\ =1.70\text{ V}-0.767\text{ V}\\ =0.933\text{ V}\end{array}$

Expression to calculate value of $K$ is as follows:

  $K={10}^{n{E}^{\text{o}}\text{/0}\text{.05916}}$        (1)

Substitute 1 for $n$ and $0.933\text{ V}$ for $E^{\text{o}}$ in equation (1).

  $\begin{array}{c}K={10}^{\left(1\right)\left(0.933\text{ V}\right)\text{/0}\text{.05916}}\\ =6\times {10}^{15}\end{array}$

Hence value of $K$ is $6\times {10}^{15}$.