Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The following steps are unbalanced half-reactions involved in the nitrogen cycle. Balance each half-reaction to show the number of electrons lost or gained, and state whether it is an oxidation or a reduction (all occur in acidic conditions):

(a)N2(g)NO(g)

(b)N2O(g)NO2(g)

(c)NH3(aq)NO2-(aq)

(d)NO3-(aq)NO2-(aq)

(e)N2(g)NO3-(aq)

Short Answer

Expert verified

a. Oxidation, N2(g)+O2(g)2NO(g)+4e-.

Gaining oxygen is oxidation.

b. Oxidation, 2N2O+3O24NO2+12e-.

Gaining oxygen is oxidation.

c.Oxidation, 2NH4+(aq)+3O22NO2-(aq)+4H+(aq)+2H2O(l)

Losing hydrogen is oxidation.

d. Reduction, 2NO3-(aq)2NO2-(aq)+O2(g)

Losing oxygen is oxidation.

e. Oxidation: 2N2(g)+5O2(g)+2H2O(l)4NO3-(aq)+4H+(aq)

Losing hydrogen is oxidation.

Step by step solution

01

Oxidation

Oxidation is a chemical reaction that results in the loss of one or more electrons in an atom or molecule. Learn the difference between oxidation and reduction by diving further into the definition, process, and real-world examples of oxidation reactions.

02

Subpart (a)

Reduction is the process of obtaining electrons, whereas oxidation is the process of losing electrons.

The complete equation is:

N2(g)+O2(g)2NO(g)

The nitrogen half-reaction is:

N2o(g)2NII+4e-

The process is called oxidation.

03

Subpart (b)

The complete equation is:

2N2O(g)+3O2(g)4NO2(g)

The nitrogen half-reaction is:

4NI(g)4NIV(g)+12e-

The process is called oxidation.

To obtain oxygen:

3O2o(g)+12e-6O2-(g)

The process is called reduction.

Therefore, oxidation process.

04

Subpart (c)

The complete equation is:

2NH4+(aq)+3O22NO2-(aq)+4H+(aq)+2H2O(l)

So, because reaction takes place in an acidic environment, the ammonia will be in the form of NH4+rather than NH3.

The nitrogen half-reaction is:

2NIII-(aq)2NIII(aq)+12e-

The process is called oxidation.

05

Subpart (d)

The complete equation is:

2NO3-(aq)2NO2-(aq)+O2(g)

The nitrogen half-reaction is:

2NV(g)+4e-2NIII

The process is called reduction.

06

Subpart (e)

The whole equation is:

2N2(g)+5O2(g)+2H2O(l)4NO3-(aq)+4H+(aq)

The nitrogen half-reaction is:

2N2o(g)4NV+20e-

The process is called oxidation.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) What is a kinetic isotope effect?

(b) Do compounds of hydrogen exhibit a relatively large or small kinetic isotope effect? Explain.

(c) Carbon compounds also exhibit a kinetic isotope effect. How do you expect it to compare in magnitude with that for hydrogen compounds? Why?

When gold ores are leached with solutions, gold forms a complex ion,Au(CN)2- , (a) Find Ecellfor the oxidation in air(PO2= 0.2l) ofAu torole="math" localid="1663334129218" Au+ in basic (pH13.55)solution with[Au+]= 0.50M Is the reactionAu+(aq) +e-Au(s) E°=1.68Vspontaneous? (b) How does formation of the complex ion change so that the oxidation can be accomplished?

Below 912°C,pure iron crystallizes in a body-centered cubic structure (ferrite) with a density ofrole="math" localid="1663336837557" 7.86g/cm3; from912°Cto1394°C, it adopts a face-centered cubic structure (austenite) with a density of7.40g/cm. Both types of iron form interstitial alloys with carbon. The maximum amount of carbon is 0.0218mass%in ferrite and 2.08mass%in austenite. Calculate the density of each alloy.

Like heavy water (D2O), so-called “semi-heavy water” (HDO) undergoes H/D exchange. The scenes below depict an initial mixture of HDO and H2reaching equilibrium.


a) Write the balanced equation for the reaction. (b) Is the value of K greater or less than? (c) If each molecule depicted represents0.10M, calculate K.

The production of S8from the H2S(g)for step 2, found in natural gas deposits occurs through the Claus process (Section 22.5):

(a) Use these two unbalanced steps to write an overall balanced equation for this process:

  1. H2S(g) +O2(g)S8(g) + SO2(g) +H2O(g)
  2. localid="1663409018375" H2S(g) + SO2(g)S8(g) +H2O(g))
    (b) Write the overall reaction with Cl2as the oxidizing agent instead ofO2.Use thermodynamic data to show whether Cl2(g)can be used to oxidizeH2S(g).
    (c) Why is oxidation bypreferred to oxidation byCl2?
See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free