Chapter 13, Problem 13.104PAE

13.106 Hydrazine, ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ , has been proposed as the fuel in a fuel cell in which oxygen is the oxidizing agent. The reactions are

   $\begin{array}{l}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\left(\text{aq}\right)+4{\text{OH}}^{\text{-}}\left(\text{aq}\right)\to {\text{N}}_{\text{2}}\left(\text{g}\right)+4{\text{H}}_{\text{2}}\text{O}\left(\mathcal{l}\right)+4{\text{e}}^{-}\\ {\text{O}}_2\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(\mathcal{l}\right)+4{\text{e}}^{-}\to 4{\text{OH}}^{\text{-}}\left(\text{aq}\right)\end{array}$

(a) Which reaction occurs at the anode and which at the cathode?

(b) What is the net cell reaction?

(C) If the cell is to produce 0.50 A of current for 50.0 h, what mass in grams of hydrazine must be present?

(d) What mass in grams of ${\text{O}}_2$ must be available to react with the mass of ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ determined in part (c)?

Interpretation Introduction

Interpretation:

To identify the net reaction.

Concept introduction:

• The net reaction is the total reaction including all reactants and products.
• It is obtained via addition of the half reactions.

Answer to Problem 13.104PAE

Solution:

Explanation of Solution

In order to find out the reaction, we must include the reduction-oxidation reactions. They always take place together, since one donates electrons and the other receives them.

In this case, the oxidation reaction: hydrazine oxidation to nitrogen gas. The reduction is oxygen forming hydroxide ion via the addition of electron

Adding both reactions, getting rid of repeating/same units on the left/right side of the equation:

Therefore, the reaction taking place is:

Interpretation Introduction

Interpretation:

To calculate the mass of hydrazine present in given amount of time and current.

Concept introduction:

• A stands for ampere, the amount of current.
• Time and current can be used to calculate total charge.
• The total charge relates mol of e- and Faraday constant.

Answer to Problem 13.104PAE

Solution:

$\text{7}\text{.46}\text{g}$

Explanation of Solution

This consist of the following steps:

1^st Step: Calculate the total amount of charge.

$\begin{array}{l}\text{t}\text{=}\text{50}\text{.0 }\text{h}\text{=}\text{50 }\text{h}\text{×}\frac{\text{3600}\text{s}}{\text{1 }\text{h}}\text{=}\text{180,000 }\text{s}\\ \text{I}\text{=}\text{0}\text{.50}\text{A}\text{=}\text{0}\text{.50}\text{C/s}\\ \text{q}\text{=}\text{I×t}\text{=}\text{0}\text{.50}\text{C/s}\text{×}\text{180,000 }\text{s}\text{=}\text{90,000}\text{C}\\ \text{1}\text{mol }{\text{e}}^{\text{-}}\text{=96500 }\text{C}\\ \text{x}\text{mol }{\text{e}}^{\text{-}}\text{=}\text{90000 }\text{C}\\ \text{x}\text{=}\text{0}\text{.9326 }\text{mol }{\text{e}}^{\text{-}}\end{array}$

2^nd Step: Calculate the total amount of moles of hydrazine

$\begin{array}{l}\text{4 }\text{mol }{\text{e}}^{\text{-}}\text{=}\text{1}\text{mol }{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\\ \text{0}\text{.9326}\text{mol }{\text{e}}^{\text{-}}\text{=}\text{y}\text{mol }{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\\ \text{y}\text{=0}\text{.2332}\text{mol }{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\end{array}$

3^rd Step: Calculate the mass of hydrazine

$\text{mass}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{=}\text{mol}\text{×}\text{Molar}\text{Mass}\text{=}\text{0}\text{.2332 }\text{mol}\text{×}\text{32}\text{.0 }\text{g/mol}\text{=}\text{7}\text{.46}\text{g}$

Interpretation Introduction

Interpretation:

To calculate the mass oxygen required

Concept introduction:

• Use stoichiometric ratios.

Answer to Problem 13.104PAE

Solution:

$\text{mass}{\text{O}}_{\text{2}}\text{=}\text{7}\text{.46}\text{g}$

Explanation of Solution

Relate the moles of oxygen and hydrazine:

1^st Step: Calculate total moles of oxygen required

The ratio is 1:1 since 1 mol of hydrazine requires 1 mol of Oxygen gas to react.

2^nd Step: Calculate the total mass of oxygen gas

$\text{mass}{\text{O}}_{\text{2}}\text{=}\text{mol}\text{×}\text{Molar}\text{Mass}\text{=}\text{0}\text{.2332 }\text{mol}\text{×}\text{32}\text{.0 }\text{g/mol}\text{=}\text{7}\text{.46}\text{g}$

The mass is exactly the same as hydrazine since they have 1:1 ratio and similar molar masses.