CH 8 chem
To find the oxidation number for chlorine, recall that the sum of the oxidation numbers equals the charge on the ion (Rule #4c).
(ox no Cl) + 4(-2) = -1 (ox no Cl) - 8 = -1 (ox no Cl) = +7 Cl = +7
Hydrogen Oxi # is Oxygens Oxi # is
+1, -2
Monatomic Oxidation # is just the charge. Ex Fe^3+ Oxi # is
+3
The black, granular material that fills a dry cell in a common flashlight (between the carbon rod and the zinc shell) is manganese(IV) oxide, MnO2. The oxidation number of manganese in MnO2 is
+4.
Oxidation
-Gain oxygen -Lose Hydrogen -Lose Electrons - Increase oxidation #
Reduction
-Lose oxygen -Gain hydrogen -Gain electrons -Decrease oxidation #
FACT -Oxidizing agents cause oxidation. Reducing agents cause reduction.
...
Elemental forms Oxi # is ___ ex; Al, Cl2, S8
0
The oxidation number for I in I2 is
0.
Identify Oxi agent and Reducing agent 1) 3 C + Fe2O3 ---> 3CO + 2Fe 2) P4 + 5O2 ----> P4O10
1) C is the REDUCING AGENT, Fe is the OXIDIZING AGENT 2) P is oxidized so its the REDUCING agent.
Which has undergone Oxidation and reduction 1) 4Cu + O2 ----> 2Cu2O 2) Mg + 2H ---> Mg ^2+ + H2
1)Cu Oxi, O Red 2) Mg oxi, H red
Oxidation numbers are assigned according to these rules:
1. A metal or a nonmetal in the free state has an oxidation number of 0. 2. A monoatomic ion has an oxidation number equal to its ionic charge. 3. H and O atoms in compounds are usually assigned oxidation numbers of +1 and -2, respectively. 4a. For a molecular compound, the more electronegative element is assigned a negative oxidation number equal to its charge as an anion. 4b. For an ionic compound, the sum of the oxidation numbers for each of the atoms in the compound is equal to 0. 4c. For a polyatomic ion, the sum of the oxidation numbers for each of the atoms in the compound is equal to the ionic charge on the polyatomic ion.
C undergone reduction or Oxi 1. C + 2Cl2 ----> CCl4 2. 2C + O2 --> 2CO
1. oxi 2. oxi
In which of the following changes is the reactant undergoing oxidation? Cu2+→Cu+ C10H16N5O14P3→C10H18N5O14P3 2Cl−→Cl2
2Cl−→Cl2
Case of Loss of Oxygen Atoms
2PbO2----> 2PbO + O2 First it has 2 O atoms attached, and then it has one
Which is the oxidizing agent in the following > Cu + 2Ag -----> Cu 2+ + 2Ag
Ag+
What is the oxidation number for barium and chloride in BaCl2?
Ba is present as Ba2+, so Ba = +2 (Rules #2 and #4b). Cl is present as Cl-, so Cl = -1 (Rules #2 and #4b).
What is reduced in the following reaction? 2 Bi3+ + 3 Mg → 2 Bi + 3 Mg2+
Bi3+
n the following change, is the reactant undergoing oxidation or reduction? (This is not a complete chemical equation.) C2H6O → C2H4O
C2H6O: Each of the six hydrogen atoms has an oxidation number of +1, thus contributing the "charge" of +6. Oxygen has an oxidation number of -2. This means that the overall "charge" due to hydrogen and oxygen is +6 − 2 = +4. The overall charge of the molecule is 0. This means that the two carbon atoms must cancel the +4 "charge." Because there are two carbon atoms, each carbon must have the oxidation number of -2. C2H4O: Each of the four hydrogen atoms has an oxidation number of +1, for a total "charge" of +4. When combined with the oxidation number of the oxygen atom equal to -2, the overall "charge" of +4 - 2 = +2. Because all molecules are neutral, each of the two carbon atoms must have the oxidation number of -1. This means that the oxidation number for carbon has changed from -2 to -1 (it became less negative in this case) and an electron was lost. In other words, carbon was reduced.
Case of Loss of Hydrogen Atoms (with compound losing H atoms)
CH3OH ------> CH2O + H2 C and O have 4 H's , then they only have 2
Case of Gain of Oxygen Atoms
CH4 + 2O2 -------> CO2 + 2H2O On CH4, C or H has no O atoms attached, but after they both have O atoms attached
Oxidation half reaction for Ca Al
Ca --> Ca^2+ + 2e- Al ----> Al3+ + 3e-
In the reaction Cu+ + Fe → Cu + Fe2+ (Note: The reaction is not balanced. whats reducing agent
Fe is the reducing agent.
Identify the oxidizing agent and the reducing agent in each reaction.3C+Fe2O3→CO+2Fe
Fe2O3 is the oxidizing agent and C is the reducing agent.
In the reaction Fe2O3 + 3 H2 → 2 Fe + 3 H2O the reducing agent is
H2.
Which substance is an oxidizing agent? HI LiBr NaOH H2O2
H2O2
Which is oxidized/reduced 2 I ^-1 + Cl2 ---> I2 + 2Cl ^-1
I is oxidized ( increase in oxi #) Cl is reduced
2H2 + O2 ------> 2H2O Explain
In this reaction, oxygen is added to hydrogen , oxidizing it. At the same time, the oxygen is reduced.
What is the oxidation number for metal, Mg?
Mg = 0 according to Rule #1.
Mg + Cl2 -------> Mg ^ +2 + 2Cl^- in terms of electrons this is
Mg oxidized bc lost electrons, Cl gained electrons so it was reduced
What are the oxidation numbers for chlorine and oxygen in the perchlorate ion, ClO4-?
O = -2 (Rule #3).
To summarize for electron gain or loss: -Oxidation -Reduction -Oxidizing agent -Reducing agent
Oxidation—the loss of electrons Reduction—the gain of electrons Oxidizing agent—the substance being reduced Reducing agent—the substance being oxidized Helpful mnemonics: OIL RIG—Oxidation Is Loss (of electrons); Reduction Is Gain (of electrons). LEO the lion says GER—Lose Electrons Oxidation; Gain Electrons Reduction.
S^-2 ------> S is
Oxidized
what gets reduced / oxidized 2Ca(s)+O2(g)→2CaO(s) F2(g)+2Li(s)→2LiF(s)
Oxo ; Ca and Li Red; O and F
F2 ---> 2F^-1 is
Reduced
What is the oxidation number for sulfur in the sulfide ion, S2-?
S = -2 (Rule #2).
What are the oxidation numbers of atoms in the following anion? d. Cr2O72-
Step 1: Place boxes above elements. Multiply each box by the number of atoms of that element in a molecule. Boxes will hold oxidation numbers. Step 2: Set up an "equation" in which the sum of the oxidation numbers of all atoms equals 0 (for a molecule) or the ion's charge (for an ion), here -2. Step 3: Follow oxidation number assignment rules for oxygen (-2). Step 4: Determine the oxidation number for chromium (+6).
What are the oxidation numbers of atoms in the following molecule? b. HNO2
Step 1: Place boxes above elements. Multiply each box by the number of atoms of that element in a molecule. Boxes will hold oxidation numbers. Step 2: Set up an "equation" in which the sum of the oxidation numbers of all atoms equals 0. Step 3: Follow oxidation number assignment rules for oxygen (-2) and hydrogen (+1). Step 4: Determine the oxidation number for nitrogen (+3).
What are the oxidation numbers of atoms in the following molecules and ions? a. SO2
Step 1: Place boxes above elements. Multiply each box by the number of atoms of that element in a molecule. Boxes will hold oxidation numbers. Step 2: Set up an "equation" in which the sum of the oxidation numbers of all atoms equals 0. Step 3: Follow oxidation number assignment rules for oxygen (-2). Step 4: Determine the oxidation number for sulfur (+4).
What happens to the oxidation number of one of its elements when a compound is oxidized?When it is reduced?
The oxidation number is increased ,The oxidation number is decreased.
t/f An oxidizing agent is reduced.
True
Identify the oxidizing and reducing agents in the following reactions. 2 C + O2 → 2 CO b. N2 + 3 H2 → 2 NH3
a. C gains oxygen and is oxidized, so it must be the reducing agent. O2 is therefore the oxidizing agent. b. N2 gains hydrogen and is reduced, so it is the oxidizing agent. H2 therefore is the reducing agent.
In each of the following, is the reactant undergoing oxidation or reduction? (These are not complete chemical equations.) a. C2H6O → C2H4O b. C2H2 → C2H6
a. oxidized b. reduced
Is the reactant undergoing oxidation or reduction? (not complete equations.) a. Zn → Zn2+ b. Fe3+ → Fe2+ c. S2− → S d. F2 → 2 F−
a. oxidized b. reduced c. oxidized d. reduced
n each of the following changes, is the reactant undergoing oxidation or reduction? (These are not complete chemical equations.) a. Pb → PbO2 b. SnO2 → SnO c. KCIO3 → KCI d. Cu2O → 2 CuO
a. oxidized b. reduced c. reduced d. oxidized
Every reduction reaction MUST be accompanied by
an oxidation reaction.
Identify the oxidizing and reducing agents in the following reactions. c. 3 Mg + N2 → Mg3N2 d. 2 H2S + 8 NO3- → SO42- + 4 N2 + 4 H2O + 2 H+
c. The oxidation number of magnesium increases from 0 to +2 as a result of magnesium losing two electrons. In other words, Mg is oxidized and is the reducing agent. As a result, N2 is the oxidizing agent, gaining electrons by decreasing the oxidation number of the nitrogen atom from 0 to -3. Note that the argument involving oxygen gain or hydrogen loss for oxidation would not have worked in this case! Neither of these elements is in this reaction. S in H2S increases its oxidation number from -2 to +6 by losing eight electrons and becoming oxidized, so H2S is the reducing agent. The oxidation number of the nitrogen atom in the NO3- ion decreases from +5 to 0: N is reduced. Consequently, NO3- is the oxidizing agent.
Most important redox reaction is ________, which is the burning of a substance in oxygen.
combustion
Reduction is the _______ of oxidation #
decrease
Whenever oxidation occurs, reduction must also occur in an exactly _________ ________, and vise versa
equillivent amount
A decrease in Oxidation # means a ______ of electrons
gain (Gain of els = reduction)
Explain for C2H4→C2H6.
gain of hydrogen atoms
oxidation number describes
how many electrons have been lost or gained by an atom.
Oxidation is the _______ of oxidation #
increase
Oxidation __________ the oxidation number , and reduction ________ the oxidation number
increases, decreases
Increase in Oxidation # means a ______ of electrons
loss ( loss of elec = oxi)
Explain for MnO2→Mn.
loss of oxygen atoms
Nitrogen dioxide converts to dinitrogen tetroxide when cooled.
neither
In a reaction, the substance undergoing reduction serves as the
oxidizing agent.
Oxidation is GAIN of ___________ or LOSS of _________ and ________
oxygen atoms , hydrogen atoms and electrons
Reduction is the LOSS of _______, or GAIN of ______ and _______
oxygen atoms, hydrogen atoms and electrons
Oxi # goes from +2 to +4, so its a ______ agent
reducing
Hydrogen gas converts tungsten oxide to tungsten metal. Hydrogen (H2) is a(n)
reducing agent.
Sodium bisulfite converts bromine (Br2) to bromide (Br-). Sodium bisulfite is a(n)
reducing agent.
A green solution of chromium(III) ions, Cr3+(aq), is converted to a blue solution of Cr2+(aq). this is
reduction
Rank the following elements in terms of their ability to act as oxidizing agents. ; Br , I , Cl, F
strongest F, Cl, Br, I Weakest
Elements in group 1&2 have oxi number of;
their charge (+1, +2)
CO + 2H2 ----------> CH3OH
this is REDUCED bc CO gains hydrogen atoms
Oxidation and reduction always occur ________
together
