Chapter 17, Problem 17.13P

(a)

Interpretation Introduction

Interpretation:

Moles of ${\text{H}}_{\text{2}}$ produced has to be calculated.

Concept Introduction:

Moles of ideal gas present in a gas container are calculated by the following equation as follows:

    $n=\frac{PV}{RT}$

Here,

$n$ is mole of ideal gas.

$P$ is pressure.

$V$ is volume.

$T$ is temperature in Kelvin scale.

$R$ is the universal gas constant.

Answer to Problem 17.13P

$1.9473\times {10}^{-3}\text{mol}$ hydrogen gas is produced.

Explanation of Solution

Formula to calculate moles is as follows:

  $n=\frac{PV}{RT}$        (1)

Substitute $\text{303}\text{K}$ for $T$, $\text{0}\text{.996}\text{bar}$ for $P$, $\text{0}\text{.082}\text{L}\text{atm}{\text{mol}}^{\text{-1}}\cdot {\text{K}}^{\text{-1}}$ for $R$ , and $\text{49}\text{.22}\text{mL}$ for $V$ in equation (1).

  $\begin{array}{c}\text{Moles}\text{of}{\text{H}}_{\text{2}}\text{gas}=\frac{\left(\text{0}\text{.996}\text{bar}\right)\left(\text{49}\text{.22}\text{mL}\right)}{\left(\text{0}\text{.082}\text{L}\text{atm}{\text{mol}}^{\text{-1}}\cdot {\text{K}}^{\text{-1}}\right)\left(\text{303}\text{K}\right)}\\ =\frac{\left(\text{0}\text{.996}\text{bar}\right)\left(\frac{\text{1}\text{atm}}{\text{1}\text{.01325}\text{bar}}\right)\left(\text{49}\text{.22}\text{mL}\right)\left(\frac{\text{1}\text{L}}{\text{1000}\text{mL}}\right)}{\left(\text{0}\text{.082}\text{L}\text{atm}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\right)\left(\text{303}\text{K}\right)}\\ =1.9473\times {10}^{-3}\text{mol}\end{array}$

Hence, number of moles is $1.9473\times {10}^{-3}\text{mol}$.

(b)

Interpretation Introduction

Interpretation:

Concentration of EDTA in molarity has to be determined.

Concept Introduction:

In analytical chemistry, coulometric titration is a method in titrating agent is produced in solution by electrolysis. Amount of that substance is determined by measuring the quantity of electricity used to electrolyze that substance. For coulometric titration, no standard solution is needed.

Moles of electron flow in a coulometric titration is calculated by the equation as follows:

    $\text{Moles}\text{of}\text{electron}\text{flow}=\frac{\left(\text{Electric}\text{current}\right)\left(\text{time}\right)}{\text{Faraday}\text{constant}}$

Answer to Problem 17.13P

Concentration of EDTA solution is $0.04112\text{M}$.

Explanation of Solution

Reduction reaction of water molecule to form hydrogen gas is written as follows:

    ${\text{2H}}_{\text{2}}\text{O}\text{+}{\text{2e}}^{-}\to 2{\text{OH}}^{-}+{\text{H}}_{\text{2}}$

Here total number of electrons flow has taken place is calculated as follows:

    $\begin{array}{c}\text{Moles}\text{of}\text{electron}\text{flow}=\left(2\right)\left(\text{Mole}\text{of}\text{hydrogen}\text{gas}\text{produced}\right)\\ =\left(2\right)\left(1.9473\times {10}^{-3}\text{mol}\right)\\ =3.8946\text{mol}\end{array}$

As EDTA is a two-electron donor, so concentration of EDTA solution can be calculated as follows:

    $\begin{array}{c}\text{Concentration}\text{of}\text{EDTA}\text{solution}=\frac{\left(\frac{3.8946\times {10}^{-3}\text{mol}}{2}\right)}{\left(\text{47}\text{.36}\text{mL}\right)\left(\frac{\text{1}\text{L}}{\text{1000}\text{mL}}\right)}\\ =0.04112\text{M}\end{array}$

Hence, concentration of EDTA solution is $0.04112\text{M}$.

(c)

Interpretation Introduction

Interpretation:

Time required for electrolysis has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 17.13P

Time taken for electrolysis is $\text{4}\text{.75}\text{hr}$.

Explanation of Solution

Formula to calculate time is as follows:

  $\text{time}\text{in}\text{second}=\frac{\left(\text{Moles}\text{of}\text{electron}\text{flow}\right)\left(\text{Faraday}\text{constant}\right)}{\left(\text{Electric}\text{current}\text{in}\text{amperes}\right)}$        (2)

Substitute $3.8946\times {10}^{-3}\text{mol}$ for moles of electron flow, $96485\text{C}\cdot {\text{mol}}^{-1}$ for faraday constant and $\text{0}\text{.02196}\text{A}$ for electric current in equation (2).

    $\begin{array}{c}\text{time}\text{in}\text{second}=\frac{\left(3.8946\times {10}^{-3}\text{mol}\right)\left(96485\text{C}\cdot {\text{mol}}^{-1}\right)}{\left(\text{0}\text{.02196}\text{A}\right)}\\ =17111.59\text{s}\\ =\text{4}\text{.75 hour}\end{array}$

Hence, time taken for electrolysis is $\text{4}\text{.75 hour}$.