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Oxidation-Reduction (a.k.a., Redox) Reactions. In redox reactions, electrons are transferred between species. Oxidation is the loss of e – s, so when a substance is oxidized, its charge… increases. LEO: “GER…”. L osing e lectrons: o xidation. G aining e lectrons: r eduction.

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slide1

Oxidation-Reduction (a.k.a., Redox) Reactions

In redox reactions, electrons

are transferred between species.

Oxidation is the loss of

e–s, so when a substance

is oxidized, its charge…

increases.

LEO: “GER…”

Losing electrons: oxidation.

Gaining electrons: reduction.

Think about what O wants to

do when it gets near stuff.

Reduction is the gain of

e–s, so when a substance

is reduced, its charge…

decreases.

“OIL RIG.”

Oxidation is loss;

reduction is gain.

slide2

Mg(s) + 2 H+(aq) Mg2+(aq) + H2(g)

0

1+

2+

2+

0

(oxidized)

(reduced)

Whenever one substance

is oxidized,

another is reduced.

“reductant”

“oxidant”

“reducing

agent”

“oxidizing

agent”

oxidation

rust, from the _________

of Fe to Fe3+

(and the reduction of O2 to O2–)

slide3

(rare)

Rules for Assigning Oxidation Numbers

1. Atoms in their elemental form have

an oxidation number of zero.

F2, Al, Mo

2. For a monatomic ion, the oxidation

number is the charge on the ion.

Ca2+, Pb4+, N3–

3. Nonmetals can have variable oxidation numbers.

a. Oxygen is usually 2–, but in the peroxide ion

(O22–) it is 1–.

b. Hydrogen is 1+ when bonded to nonmetals,

1– when bonded to metals.

** H– can happen only with the metals.

c. Fluorine is 1–. Other halogens are usually 1–, but are +

when combined w/oxygen.

in BrO2–, Br is… 3+

4. The sum of the oxidation numbers in a

neutral compound is zero.

H2O, Fe2O3

slide4

Determine the oxidation number

of nitrogen in each of the following.

N2O4

NH3

4+

3–

N2O4 is a key

component of

smog.

NH3 is used

in many types

of cleaners.

N2

NO3–

0

5+

N2 makes up nearly 80%

of Earth’s atmosphere.

Potassium nitrate (saltpeter) is

used in the making of black powder.

slide5

Single replacement reactions have the following form:

A + BX  AX + B

Write molecular and net ionic equations for the

calcium/hydrochloric acidreaction.

Ca(s) + 2 HCl(aq) CaCl2(aq) + H2(g)

Ca(s) + 2 H+(aq) Ca2+(aq) + H2(g)

slide6

easily lose e–

don’t like losing e–

Li

Rb

K

Ba

Sr

Ca

Na

Mg

Al

Mn

Zn

Cr

Fe

Cd

Co

Ni

Sn

Pb

H2

Sb

Bi

Cu

Hg

Ag

Pt

Au

The activity series is a list of metals.

At the top of the list

are the highly-

reactive active metals.

At the bottom are the

not-so-reactive noble metals.

The activity series is readily available

in standard references.

(i.e., …will change Pb2+ to Pb?)

Which of the following metals

will reduce PbCO3? Ag Mg Hg

slide7

Molarity

mol

A solution’s concentration tells us

the amount of solute per solvent.

M

L

A common unit of concentration

is molarity.

-- equation:

What mass of magnesium nitrite is needed to make

3.25 L of a 0.35 M solution?

Mg2+

NO2–

Mg(NO2)2

mol = M L

= 0.35 M (3.25 L )

= 1.1375 mol

1.1375 mol

= 130 g Mg(NO2)2

slide8

Steps for Properly Mixing an Aqueous Solution

1. Fill an appropriate container (e.g., graduated

cylinder or volumetric flask) mostly full of water

(~80% full). This is an approximate technique

and should take very little time.

2. Weigh out the proper amount

of solute and mix it into the

water from Step 1.

3. “Top off” the solution to the

proper volume and mix.

DONE.

slide9

strong electrolyte

What is the conc. of sodium ions in a 0.025 M

solution of sodium phosphate?

Na+

PO43–

Na3PO4

Na3PO4(aq)

3 Na+(aq) + PO43–(aq)

0.025 M

0.075 M

slide10

Dilutions

Aqueous acids (and sometimes bases) can be

purchased in concentrated form and diluted to any

lower concentration. A purchased bottle of acid is

called a concentrate or a stock solution.

-- **Safety Tip: When diluting, add acid or base

to water, not the other way around.

C = conc.

D = dilute

Dilution Equation:

MCVC = MDVD

Conc. phosphoric acid is 14.8 M. What volume of

concentrate is req’d to make 25.0 L of 0.500 M acid?

(VC)

=

14.8

0.500

(25.0)

VC = 0.845

L

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