Changes in oxidation states are identified in this method.
A loss of five electrons per N atom is the result of a nitrogen increase from -2 in NH3 to +2 in NO.
Oxygen decreases from 0 in O2 to -2 in NO and H2O, corresponding to a "gain" of two electrons per O atom.
The gain of 10 e and the loss of are related to the proportion of N to O atoms.
The meaning of these terms is briefly discussed.
These species can't be reduced in redox reactions.
The oxidizing agent is reduced.
The reducing agent oxidizes itself in the reaction.
A substance with an element in one of its highest possible oxida tion states is an oxidizing agent.
The substance is a reducing agent if it is in one of the lowest oxidation states.
The oxidation states of nitrogen are shown in Figure 5-16.
The oxidation state of the nitrogen in dinitrogen tetroxide 1N2O42 is nearly the maximum value that can be achieved.
The nitrogen atom in hydrazine 1N2H42 is in the lowest oxidation state and therefore it is a reducing agent.
The reducing agent is N2H4 while the oxidizing agent is N2O4.
The reaction releases a lot of energy.
Certain substances in which the oxidation state of an element is between its highest and lowest possible values may act as oxidizing agents in some instances and reducing agents in others.
hydrazine acts as an oxidizer when it is combined with hydrogen to make ammonia.
There are many uses for the oxidizing agent, marangganate ion, in the chemical laboratory.
The use of quantitative analysis of iron is described in the next section.
Ozone, O31g2, a triatomic form of oxygen, is an oxidizer used in water purification.
6 CO21g2 + 3 H2O1l2 + 14 O21g2 is a powerful oxidizer.
It is used in liquid chlorine bleaches.
The reduction of the OCl- ion to Cl- is associated with the bleaching action of NaOCl.
A red cloth becomes white when immersed in NaOCl, metabolism, and the transport of oxygen.
H2O2 is a versatile chemical.
It can be used in water purification and as a substitute for chlorine.
It can be either an oxidizing or a reducing agent.
Determine if hydrogen peroxide is an oxidizing or reducing agent for the following reactions.
We need to assign oxidation states and identify which substance is being reduced before we can identify the reducing agents.
Another substance is oxidation caused by the oxidizing agent.
Another substance is reduced by the reducing agent.
The oxidation state of oxygen in H2O2 is -1.
It is -2 in H2O.
The reduction of hydrogen peroxide acts as an oxidizer.
The reducing agent is hydrogen peroxide.
Hydrogen peroxide acts as a reducing agent.
H2O2 can act as an oxidizing agent and a reducing agent.
H2O2 is reduced to an acidic solution when it acts as an oxidizing agent.
It oxidizes to O21g2 when it acts as a reducing agent.
In the following reaction, identify the oxidizing and reducing agents.
A newspaper account of an accidental spill of hydrochloric acid in an area where sodium hydroxide solution was also stored spoke of the potential hazardous release of chlorine gas if the two solutions came into contact.
If our goal is to get the maximum yield of a product at the lowest cost, we would generally choose the most expensive reactant as the limiting reactant and use excess amounts of the other reactants.
In most precipitation reactions, this is the case.
We may not be interested in the products of a reaction but in the relationship between two reactants in some instances.
We have to carry out the reaction in a way that neither reactant is excessive.
A solution of one reactant is placed in a small beaker.
The stopcock is manipulated to add the second solution.
A small quantity of water, a few drops of phenolphthalein indicator, and a sample of vinegar are added to a flask.
The solution becomes basic when an additional drop of NaOH(aq) is added.
The indicator is pink.
The first appearance of the pink color is considered to be the equivalent of the titration.
The key to every titration is that at the equivalence point, the two reactants have been consumed and neither remains in excess.
The key to a successful signal is when the equivalence point is reached.
A small amount of the right indicator.
When we are near the equivalence point, we add the indicator to the reaction mixture to learn how to make a color change.
titration data is used in some calculations.
We usually can get iron wire in pure form and allow it to react with three or four significant acids to yield Fe2+1aq2.
Fe2 is converted to Fe3 in figures.
An acidic solution is not important.
Determine the volume of KMnO41aq2 required to oxidize a known quantity of Fe2+ round number.
We can calculate the exact molarity 0.1035 M, but not the KMnO41aq2.
The buret is filled with intensely colored KMnO41aq2 and the solution contains a known amount of Fe2+.
The pink color is caused by a fraction of a drop of the KMnO41aq2 beyond the equivalence point.
A solution of acetic acid produced by the fermentation of apple cider, wine, or other carbohydrate material can be used to establish the concentration of acids and bases.
The minimum acetic acid content is 4%.
A sample of a particular substance is measured with a large amount of NaOH.
CH3COOH is a weak acid and NaOH is a strong base.
The acid-base neutralization reaction is between CH3COOH and NaOH.
A balanced chemical equation is written for the reaction.
We need to convert mL NaOH to g CH3COOH.
The equation for the reaction is given below.
The mass of CH3COOH has a density of 1.01 g>mL.
The sample is slightly above the legal minimum limit.
The maximum amount of acetic acid is allowed.
This titration technique can be used to make sure that the vinegar stays between the limits.
This is a problem that involves many steps or conversions.
Try to break the problem into simpler ones.
It's important to remember that there are three steps to solve a stoichiometry problem: (1) converting to moles, (2) converting between moles and (3) converting from moles.
To carry out mole-mole conversions and gram-mole conversions, use molarities and molar mass.
A balanced chemical equation is used to construct the stoichiometric factors.
A sample of KHC8H4O4 is dissolved in water and then measured with 24.03 mL of NaOH.
A 0.235 g sample of a solid that is 92.5% NaOH and 7.5% Ca1OH22 needs 45.6% of a HCl(aq) solution for its titration.
A sample of 0.211 M KOH is added to a sample of 0.
100 M HCl.
A piece of iron wire weighing 0.1568 g requires a KMnO41aq2 solution to be used for its titration.
The amount of two reactants that are consumed in the titration are equivalent.
We must determine the number of moles of KMnO4 in the 26.24 mL sample by using a mass of Fe.
Determine the amount of KMnO4 consumed in the titration.
We use solutions that are not very large or small for practical applications.
If you calculate a molarity that is larger than 0.1 M or smaller than 0.001 M, you must check your calculation for possible errors.
A sample of iron is dissolved in acid and the iron is reduced to Fe2 and then taken to a lab for testing.
Determine the mass percent Fe.
sodium oxalate is a substance that may be used to standardize.
Most of us don't take for granted access to a plentiful supply of pure water.
Most of us agree that water purification is important.
The way in which water is treated depends on how it is used.
For a discussion of the removal or destruction of undesirable chemical substances from water, go to the Focus On feature for Chapter 5, entitled Water Treatment, on the MasteringChemistry site.
Break of the individual solute species present-- down the reaction into separate half-reactions-- is an effective way to balance a redox equation.
Titration data can be used to establish a product.
Na2S2O4 is an important reducing agent.
Another product is sulfate ion.
The chro mate ion can be found in wastewater from a plant.
White solid sodium dithionite, Na2S2O4, is added to a yellow solution.
The product of the reaction is gray-green.
Carey B. V is a reaction.
To get a balanced chemical equation for the reaction, we need to convert 100.0 L of wastewater into grams of Na2S2O4.
An ionic expression is used to represent a reaction.
If the half-reactions occur in acidic solution, there will be 3 + H2O.
The conditions should be changed to a basic solution.
2 H2O and 2 OH 2 H+ to form 2 H2O on the right.
The main effort was to balance the equation for the reaction under basic conditions.
We were able to find the relationship between dithionite and chromate ion.
The rest of the problem was a calculation for a reaction in solution.
A sample of 0.1432 g was used to determine the amount of potassium chlorate.
The solution was acidified after the sample was dissolved.
The back-titrated solution was 0.08362 M Ce1NO324.