Oxidation-Reduction Reactions

Oxidation-Reduction Reactions
Chapter 5 Oxidation-Reduction Reactions

Oxidation Reactions - -principal source of energy on earth - -combustion of gasoline - -burning of wood - digestion of food in your body

Contents 5.1. Oxidation-reduction reactions involve electron transfer 5.2. The ion-electron method creates balanced net ionic equations for redox reactions 5.3. Metals are oxidized when they react with acids 5.4. A more active metal will displace a less active one from its compounds 5.5. Molecular oxygen is a powerful oxidizing agent 5.6. Redox reactions follow the same stoichiometric principles as other reactions

Oxidation reactions are always accompanied by a reduction reaction
- originally meant combining with oxygen - iron rusting (iron + oxygen)

Oxidation reactions are always accompanied by a reduction reaction
Reduction - originally meant the loss of oxygen from a compound - removing iron from iron ore ( iron II oxide) e.g. smelting

Ca(s) + 2 H2O(l) → Ca(OH)2(s) + H2(g)
Oxidation - Reduction Involve two processes: Oxidation – the loss of electrons, and Reduction – the gain of electrons Ca(s) H2O(l) → Ca(OH)2(s) + H2(g) Oxidizer = Oxidizing agent = received the electrons - substance being reduced Reducer = Reducing agent = donated the electrons - substance being oxidized

Reduction is… Oxidation is… the gain of electrons
a decrease in oxidation state the loss of oxygen the addition of hydrogen MgO + H2 ® Mg + H2O notice the Mg2+ in MgO is gaining electrons Oxidation is… the loss of electrons an increase in oxidation state the addition of oxygen the loss of hydrogen 2 Mg + O2 ® 2 MgO notice the magnesium is losing electrons

Oxidation – Reduction Reactions
LEO goes GER Loss of Electrons Oxidation Gain of Electrons Reduction

QUESTION In a redox reaction, oxidation and reduction must both occur. Which statement provides an accurate premise of redox chemistry? The substance that is oxidized must be the oxidizing agent. The substance that is oxidized must gain electrons. The substance that is oxidized must have a higher oxidation number afterwards. The substance that is oxidized must combine with oxygen.

Guidelines for Redox Reactions
Oxidation and reduction always occur together Total number of e- lost by one substance is the same as the total number of e- gained by the other Conservation of mass and charge For a redox reaction to occur, something must accept the e- that are lost by another substance

Oxidation numbers Oxidation number: - of any free element is zero Free element?? - of any simple, monatomic ion is equal to the charge on the ion - of fluorine in its compounds is –1 - of hydrogen in its compounds is +1 - of oxygen in its compounds is -2

Oxidation Numbers The sum of all oxidation numbers of the atoms in a molecule or polyatomic ion must equal the charge on the particle!

Other Situations If there is a conflict between two rules apply the rule with the lower number and ignore the conflicting rule In binary ionic compounds with metals, the nonmetals have oxidation numbers equal to the charges on their anions

Assigning Oxidation Numbers
Oxidation States Oxidation states are numbers assigned to atoms that reflect the net charge an atom would have if the electrons in the chemical bonds involving that atom were assigned to the more electronegative atoms. Oxidation states can be thought of as “imaginary” charges. They are assigned according to the following set of rules:

#1) The oxidation number of a simple ion is equal to its ionic charge
Na + Cu 2+ N3-

#2) The oxidation number of hydrogen is always +1, except in metal hydrides like NaH where it is –1
HCl NaH

#3) The oxidation number of oxygen is always –2 except in peroxides like X2O2 where it is –1.
H2O H2O2

#4) The oxidation number of an uncombined (free) element is always zero
Na Cu N2

#5) For any neutral (zero charge) compound,
#5) For any neutral (zero charge) compound, the sum of the oxidation number’s is always zero +4 -2 CO2

#6) For a complex or polyatomic ion, the sum of the oxidation number’s equals the charge of the complex ion +7 -2 MnO41-

Examples - assigning oxidation numbers
Assign oxidation states to all elements:

Assign Oxidation States to All Atoms
Fe2O3 Na2CO3 V(OH)3 K2Cr2O7

Redox may be defined by Oxidation is an increase in oxidation numbers - electrons are being lost in order to increase positive charge character Reduction is a decrease in oxidation number - electrons are being gained in order to decrease positive charge character

+2 to +4 0 to -1 Look for Oxidation # Changes
an increase in oxidation number of an atom signifies oxidation +2 to +4 a decrease in oxidation number of an atom signifies reduction 0 to -1

Identifying Redox Reactions
Oxidation and reduction always occur together in a chemical reaction. For this reason, these reactions are called “redox” reactions. Although there are different ways of identifying a redox reaction, the best is to look for a change in oxidation state:

SnCl2 + PbCl4 SnCl4 + PbCl2 CuS + H+ + NO3- Cu+2 + S + NO + H2O
+2 = LEO OA +2 -1 +4 -1 +4 -1 +2 -1 SnCl PbCl SnCl PbCl2 RA -2 = GER -3 = GER RA +2 -2 +1 +5 -2 +2 +2 -2 +1 -2 CuS H NO Cu S NO H2O OA +2 = LEO

Examples - labeling redox reactions
In each reaction, look for changes in oxidation state. If changes occur, identify the substance being reduced, and the substance being oxidized. Identify the oxidizing agent and the reducing agent. = +1 (H is oxidized) (reducing agent) +2 -2 +1 -2 H2 + CuO ® Cu + H2O = -2 (Cu is reduced) (oxidizing agent)

5 Fe2+ + MnO4- + 8 H+ ® 5 Fe3+ + Mn2+ + 4 H2O
Try These!! +1 = Fe 2+ is oxidized (reducing agent) 5 Fe2+ + MnO H+ ® 5 Fe3+ + Mn H2O Zn HCl ® ZnCl2 + H2 - 5 = Mn 7+ is reduced (oxidizing agent) +2 = Zn 0 is oxidized (reducing agent) - 1 = H 1+ is reduced (oxidizing agent)

Balancing Redox Reactions – Ion – Electron Method
Identify the half-reactions Balance each atom in the half reaction, saving H and O for last Balance O by adding 1 water molecule for each needed O Balance H by adding 1 H+ ion for each needed H

Balancing Redox Reactions – Ion – Electron Method
Balance charges by adding electrons to the more positive side Find the least common multiple of electrons for the two half- reactions. Multiply each reaction by the factor needed to achieve the LCM of electrons Add the half reactions, canceling like substances that appear on both sides

Balancing Basic Reactions
The simplest way to balance reactions in basic solution is to first balance them as if they were in acidic solution, then “convert” to basic solution Additional Steps for Basic Solutions 8) To both sides of the equation, add the same number of OH- ions as there are H+. 9) Combine H+ and OH- to form H2O 10) Cancel H2O molecules that are on both sides of the reaction.

Example MnO4- + C2O42- → MnO2 + CO32-

Half Reactions Oxidation C2O > CO32- Reduction MnO > MnO2

Balance Half Reactions excepting H & O
Oxidation C2O > 2 CO32- Reduction MnO > MnO2

Balance O MnO > MnO2 + 2 H2O C2O H2O > 2 CO32-

Balance H MnO H > MnO2 + 2 H2O C2O H2O > 2 CO H+

Balance Charge MnO H+ + 3 e > MnO2 + 2 H2O C2O H2O > 2 CO H+ + 2 e-

Balance Electrons using LCM
(MnO H+ + 3 e > MnO2 + 2 H2O) x 2 (C2O H2O > 2 CO H+ + 2 e-) x 3

Add Reactants and Products
2 MnO H+ + 6 e- + 3 C2O H2O -----> 2 MnO2 + 4 H2O + 6 CO H+ + 6 e-

Cancel Where Possible 2 MnO H+ + 6 e- + 3 C2O H2O -----> 2 H2O 2 MnO2 + 4 H2O + 6 CO H+ + 6 e- 4 H+

Example cont’d 2 MnO C2O H2O > 2 MnO2 + 6 CO H+

Balance using the Ion- Electron Method
Mn(s) + Cu2+(aq) → Cu(s) + MnO2(s) (basic) Identify Half-reactions Mn > MnO2(s) Cu > Cu(s)

Add water for oxygen Mn + 2 H2O > MnO2(s) Cu > Cu(s) Add protons for hydrogen Mn + 2 H2O > MnO2 + 4 H+ Cu(s) > Cu+2

Add electrons for charge Mn + 2 H2O > MnO2 + 4 H+ + 4 e- Cu e > Cu(s) Multiply to equalize number of electrons {Mn + 2 H2O > MnO2 + 4 H+ + 4 e-} x 1 {Cu e > Cu(s) } x 2

Add and Cancel like terms Mn + 2 H2O > MnO2 + 4 H+ + 4 e- 2 Cu e > 2 Cu(s) Mn + 2 H2O + 2 Cu+2(aq) > MnO2 + 4 H+ + 2 Cu(s)

Add a OH- for each H+ in the equation Mn + 2 H2O + 2 Cu+2(aq) + 4 OH-(aq) > MnO2 + 4 H+ + 4 OH-(aq) + 2 Cu(s) Combine H+ and OH- forming water MnO2 + 4 H2O(l) + 2 Cu(s)

Cancel waters if possible. Mn + 2 H2O + 2 Cu+2(aq) + 4 OH-(aq) > MnO H2O(l) + 2 Cu(s) Result Mn + 2 Cu+2(aq) + 4 OH-(aq) > MnO2 + 2 H2O(aq) + 2 Cu(s)

ClO- + VO3-(aq) > ClO3-(aq)+ V(OH)3(s) (basic)

Your turn! What is the balanced oxidation – reduction equation for: Acid MnO4-(aq) + Fe+2(aq) > Fe3+(aq) + Mn2+(aq)

Balancing Redox Equations in acidic solutions
1) Determine the oxidation numbers of atoms in both reactants and products. 2) Identify and select out those which change oxidation number (“redox” atoms) into separate “half reactions”. 3) Balance the “redox” atoms and charges (electron gain and loss must equal!). 4) In acidic reactions balance oxygen with water then hydrogen from water with acid proton(s).

Fe 3+(aq) + Cr 3+(aq) + H2O(l)
Balancing Redox Equations Fe+2(aq)+ Cr2O72-(aq) +H+(aq)  Fe3+(aq) + Cr3+(aq) + H2O(l) Fe 2+(aq)+ Cr2O72-(aq) +H+(aq)  Fe 3+(aq) + Cr 3+(aq) + H2O(l) ? Cr oxidation number? x = ? Cr ; 2x+7(-2) = -2; x = +6

Balancing Redox Equations
Fe 2+(aq)  Fe 3+(aq) + e- Cr2O72-(aq)  Cr 3+(aq) Cr = (6+) 2 6 e - 6 (Fe 2+(aq) Fe3+(aq)) + e - 6 Fe 2+(aq)  6 Fe3+(aq) + 6 e - Cr2O72-(aq) + 6 e -  2 Cr3+(aq)

6 Fe2+(aq)  6 Fe3+(aq) + 6 e - Cr2O72-(aq) + 6 e -  2 Cr3+(aq) 6 Fe2+(aq)+ Cr2O72-(aq) + ? 2nd H+(aq)  6 Fe3+(aq) + 2 Cr3+(aq)+ ? 1st Oxygen H2O(l) Oxygen = 7 2nd (Hydrogen) = 14

Completely Balanced Equation: 6 Fe2+(aq)+ Cr2O72-(aq) H+(aq)  6 Fe3+(aq) + 2 Cr3+(aq)+ 7 H2O(l)

Cr2O72−(aq) + C2H5OH(aq) → Cr3+(aq) + CO2(g) + H2O(l)
QUESTION Dichromate ion in acidic medium converts ethanol, C2H5OH, to CO2 according to the unbalanced equation: Cr2O72−(aq) + C2H5OH(aq) → Cr3+(aq) + CO2(g) + H2O(l) The coefficient for H+ in the balanced equation using smallest integer coefficients is: A) 8 B) 10 C) 13 D) 16

Balancing Redox Equations in basic solutions
1) Determine oxidation numbers of atoms in Reactants and Products 2) Identify and select out those which change oxidation number into separate “half reactions” 3) Balance redox atoms and charges (electron gain and loss must equal!) 4) In basic reactions balance the Oxygen with hydroxide then Hydrogen from hydroxide with water

MnO2 (aq)+ ClO31-(aq) + OH 1-aq)  MnO41- (aq)+ Cl 1-(aq) + H2O(l) Mn4+ (MnO2)  Mn7+ (MnO4 ) 1- Cl+5 (ClO3 ) e-  Cl 1-

Electronically Balanced Equation: 2 MnO2 (aq)+ ClO31-(aq) + 6 e -  2 MnO Cl e-

2 MnO4 (aq)1- + Cl 1- (aq)+ 1 H2O (l)
Balancing Redox Equations in basic solutions Completely Balanced Equation: 2 MnO2 (aq)+ ClO31-(aq) + 2 OH 1- (aq)  2 MnO4 (aq)1- + Cl 1- (aq)+ 1 H2O (l) 9 O in product

MnO4– + C2O42–  MnO2 + CO32– (basic solution)
QUESTION Oxalate ion can be found in rhubarb and spinach (among other green leafy plants). The following unbalanced equation carried out in a basic solution, shows how MnO4– could be used to analyze samples for oxalate. MnO4– + C2O42–  MnO2 + CO32– (basic solution) When properly balanced, how many OH– are present? 1 2 3 4

MAP of Balancing Redox

Reactions of Metal in Oxidizing/Non-oxidizing Acids
Hydrogen can be replaced in an acid by a metal when it reacts with acid M(s) + HA(aq) > MA(aq) + H2↑ In this case, H+ is reduced so that the metal can be oxidized Recognize, hydrogen ion (H3O+) itself, however, is not a strong oxidizing agent

Only metals more active than H can undergo this single-replacement or single displacement type reaction: Au(s) + H+(aq) → N.R. Fe(s) + 2 H+(aq) →Fe2+(aq) + H2(g) Note: the product here is hydrogen gas

Nonoxidizing acids have a poor oxidizing ability. In these acids, the anion of the acid is a weaker oxidizer than the hydrogen cation (H3O+). This anion is more difficult to reduce than H+(aq).

However, some acids contain anions that are stronger oxidizers than H+(aq). These are usually considered to be oxidizing acids These acids and their oxidizing strength may depend upon conditions, concentration, and the reducing strength of the metal.

Oxidizing Acids Can React w/Most Metals
Nitric acid is considered an oxidizing acid because the nitrate ion is a better oxidizing agent than H+(aq). the products of the reduction may change pending upon concentration of the acid; and strength of the reducing agent. - (conc) NO H+(aq) + e- → NO2(g)+ H2O(l) - (dil) NO3 -(aq) + 4 H+(aq)+ 3 e- → NO(g) + 4 H2O(l) - (v.dil): NO3-(aq) + 10 H+ + 8 e- → NH4(aq)+ + 3 H2O(l)

Oxidizing Acids Can React w/Most Metals
While sulfuric acid is considered a non-oxidizing acid, when it is hot and concentrated it may become a fairly strong oxidizer. (hot, conc.) SO H+(aq) + 3 e- → SO2(g) + 2 H2O(l) - (hot, conc, with strong reducing agent) - SO4(aq) H+(aq) + 8 e- → H2S(g) + 4 H2O(l)

Reactions with Dilute Nitric Acid
3 Cu(s) + 8 H+(aq) + 2 NO3-(aq) > 3 Cu+2(aq) + 2 NO(g) + 4 H2O(l) Note: Instead of forming hydrogen gas (no bubbling observed), the product of the reduction is water.

Important Note Nitric acid may react with organic compounds in a vigorous if not explosive manner.

Single Displacement or Replacement Reactions
General Form: A + BC → AC + B Metal A can replace metal B if it is a more active metal, or General Form: A + BC → BA + C Nonmetal A can replace nonmetal C if it is more active than C. An activity series arranges metals according to their ease of oxidation (Table 5.3)

Learning Check: Metal Activity
Using the following observations, rank these metals from most reactive to least reactive Cu(s) + HCl(aq) → no reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) Mg(s) + ZnCl2(aq) → MgCl2(aq) + Zn(s) Mg > Zn > H > Cu

Activity Series

Learning Check: Predict the Products of the Following: Zn + CuSO4→
Cu + ZnSO4 → AgNO3(aq) + Cu(s) → Solid magnesium and aqueous iron(III) chloride Solid nickel and aqueous sodium chloride

Oxygen Reacts with Many Substances
Oxygen is a strong oxidizing agent It will react with many substances such as organic compounds - organic compounds?

The products depend, in part, on how much oxygen is available Combustion of hydrocarbons – these are? - O2 plentiful; CH4 + 2O2→ CO2 + 2H2O - O2 limited: 2CH4 + 3O2 → 2CO + 4H2O - O2 scant: CH4 + O2 → C + 2H2O

In the last reaction, the unburnt carbon is referred to as “soot” or industrially as “lampblack” What are some industrial applications of lampblack? Inks and dyes Tire manufacturing

Organic compounds containing O also produce carbon dioxide and water - C2H5OH + 3O2 → 2CO2 + 3H2O Organic compounds containing S produce sulfur dioxide - 2 C2H5SH + 9 O2 → 4 CO2 + 6 H2O + 2 SO2

Some organic compounds contain sulfur, in which case some of the products may be sulfur dioxide, SO2, a major source of pollution. What type?

Other Reactions with Oxygen
Many metals corrode or tarnish when exposed to oxygen - 4 F e + 3 O2 →2 Fe2O3 - 4 Ag + O2 →2 Ag2O Most nonmetals react with oxygen directly - Plentiful: C + O2 →CO2 - Limited: 2 C + O2 → 2 CO

Learning Check: Complete the Following Reactions
Aluminum metal and oxygen gas forms aluminum oxide solid. Solid sulfur (S8) burns in oxygen gas to make gaseous sulfur trioxide All ionic compounds are solids at room temperature. 5.5 Molecular oxygen is a powerful oxidizing agent

Learning Check: Complete the Following Reactions
Copper metal is heated in oxygen to form black copper(II) oxide solid.

Rare Redox Reaction Disproportionation reactions are a rare form of oxidation reduction reaction in which substances can be both oxidized and reduced. An example would be the decomposition of hydrogen peroxide. 2 H2O2(l) > 2 H2O(l) + O2(g)

Disproportionation In the example of hydrogen peroxide, oxygen is -1 in the peroxide. In forming water, oxygen is reduced, becoming -2. In forming diatomic oxygen, the oxygen is oxidized, becoming 0.

Redox Stoichiometry Calculations involving concentrations and redox reactions are quite common. Many ores containing metals are analyzed using redox titrations. Since many compounds change color as they are oxidized or reduced, one of the reactants may serve as the indicator in the titration.

Example Problem A g sample of tin ore was dissolved in acid solution converting all the tin to tin(II) ion. In a titration, 8.08 mL of M KMnO4 was required to oxidize the tin(II) to tin(IV). What was the percentage tin in the original sample?

Example cont’d Solution:

Example cont’d % tin = mass tin x 100 mass ore

Redox Stoichiometry The concentration of iron(II) can be determined by titration with bromate ion, in acid. The products are iron(III) ion and the bromide ion. What is the concentration of iron(II) ion if mL of M potassium bromate is required to completely react with mL of the iron solution.

Redox Stoichiometry The concentration of iron(II) can be determined by titration with bromate ion, in acid. The products are iron(III) ion and the bromide ion. 1. Write the balanced chemical reaction.

Redox Stoichiometry

Redox Stoichiometry M = moles solute VL sol’n

Redox Stoichiometry What volume of M permanganate solution must be used to titrate 1.55 g of ferrous ion. Recall, permanganate is a good OXIDANT!! This is run in ACIDIC media.

Redox Stoichiometry

Oxidation-Reduction Reactions

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