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Oxidation Reduction Reactions Very few metals exist in pure form in nature (exceptions are: ). Most exist as compounds with various other elements. These are called From metallurgy (the process of producing metals from their compounds or refining) this process became known as reduction. Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g) SnO2(s) + C(s) Sn(s) + CO2(g) CuS(s) + H2(g) Cu(s) + H2S(g) In each case the metal ore is ‘reduced’ to its elemental state. The substance used to cause or promote the reduction of a metal compound to an elemental metal is called the reducing agent. In the examples above the reducing agents are: The reverse of this process is called oxidation, in which a metal is reacted with another substance to form a compound. Oxidation reactions can be further broken down into two sub-categories known as combustion (burning) and corrosion (rusting). 2Mg(s) + O2(g) 2MgO(s) 2Al(s) + 3Cl2(g) 2AlCl3(s) Cu(s) + Br2(g) CuBr2(s) The substance used to cause or promote the oxidation of a metal to produce a metal compound is called the oxidizing agent. In these examples the oxidizing agents are: 000_Electric Cells Student Notes SS.doc 1 of 20 Homework: Page 653 #2,3,4 Oxidation/reduction reactions can be explained by examining the transfer of electrons between the atoms involved in the reaction. Example: Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) Write the ionic equation for the reaction: The chlorine ions are not involved in the reaction (they are ). Isolate the ‘real reactants’ and write half-reaction equations showing the transfer of electrons: For half-reactions the number of atoms on each side must be balanced (as always) and the total charge must also be balanced on each side as well. In other words, From these half-reactions it can be seen that Zn(s) loses electrons. The Zn(s) has been oxidized. The H+(aq) ions have gained electrons and have been reduced. Homework: pg 656, #8 11 000_Electric Cells Student Notes SS.doc 2 of 20 Oxidation States The oxidation state of an atom in a molecule/ion is An oxidation number Oxidation numbers are written with the +/- sign in front of the number. (i.e. +2, -1) whereas valence numbers (the actual charge on an atom/molecule/ion) are written with the +/- sign behind the number (i.e 2+, 1-). Oxidation numbers and valence numbers are not necessarily the same!!! Oxidation numbers are a contrived, arbitrary method to count electrons. Common Oxidation Numbers Atom or Ion Oxidation Number Examples All atoms in elements Hydrogen in all compounds Except hydrogen in hydrides Oxygen in all compounds Except oxygen in peroxides All monatomic ions Using these ‘known’ oxidation numbers, the oxidation numbers for any other element in a compound/ion can now be determined. 000_Electric Cells Student Notes SS.doc 3 of 20 Examples 1. What is the oxidation number of carbon in methane (CH4)? Hydrogen has an oxidation number of There are 4 hydrogens, so the total oxidation number for the hydrogen in this molecule is. The molecule is electrically neutral (it does not have a net charge), so therefore the carbon must have an oxidation number of 2. What is the oxidation number of manganese in the permanganate ion (MnO4-)? Oxygen has an oxidation number of There are 4 oxygen atoms, so the total oxidation number for the oxygen in this molecule is The ion has a net charge of 1-, so therefore the manganese must have an oxidation number of 3. What is the oxidation number of sulfur in sodium sulfate? Using the concept of oxidation numbers, the terms oxidation and reduction can be further refined to mean: Oxidation – Reduction – 000_Electric Cells Student Notes SS.doc 4 of 20 Example The determination of blood alcohol content from a sample of breath or blood involves the reaction of the sample with acidic potassium dichromate solution. If ethanol is present, chromium(III) ions, water, and acetic acid are produced. Identify the oxidation and reduction components in the following reaction. 2Cr2O72-(aq) + 16H+(aq) + 3C2H5OH(aq) 4Cr3+(aq) + 11H2O(g) + 3CH3COOH(aq) In biology, oxidation and reduction are often defined in terms of the addition or removal of oxygen or hydrogen in relation to carbon. The removal of oxygen results in the decrease of the oxidation number of carbon. HCOOH HCHO C is +2 C is 0 carbon has been reduced The removal of hydrogen results in the increase of the oxidation number of carbon. C2H6 C2H4 C is -3 C is -2 carbon has been oxidized Photosynthesis involves a series of reduction reactions where the oxidation number of carbon is gradually changed from +4 in carbon dioxide to 0 in sugars. Cellular respiration reverses this process by gradually oxidizing sugars until carbon is again +4 in carbon dioxide. Homework: Page 659 #12, 13, 14, 16 Page 662 #18, 19, 20 000_Electric Cells Student Notes SS.doc 5 of 20 Balancing Redox Reactions There are two methods that can be used to balance redox reactions. One involves using oxidation numbers and the other involves balancing the two half-reactions involved in the reaction. When balancing redox reactions Method 1: Balancing Redox Reactions using Oxidation Numbers Type 1: Simple Redox Reaction in a neutral solution Balance the following redox reaction. H2S(g) + O2(g) SO2(g) + H2O(g) Step 1: Assign each atom its oxidation number and identify the ones that change. Step 2: Determine the change in electrons per molecule involved. H2S has only one sulfur atom, O2 has two oxygen atoms, Step3: Find the lowest common factor to balance the gain and loss of electrons for each reactant. In this case the lowest common multiple is 12. So the H2S must be multiplied by 2 and the O2 must be multiplied by 3. We now have the reactants balanced as follows: 000_Electric Cells Student Notes SS.doc 6 of 20 Step 4: Balance the products using the coefficients from the reactants. (Don’t change the coefficients for the reactants since they are already balanced!!) Type 2: Redox Reaction in an acidic solution Chlorate ions and iodine react in an acidic solution to produce chloride ions and iodate ions. Balance the equation for this reaction. Write the equation using the known reactants and products. Step 1: Assign each atom its oxidation number and identify the ones that change. Step 2: Determine the change in electrons per molecule involved. ClO3 has only one chlorine atom, I2 has two iodine atoms, Step 3: Find the lowest common factor to balance the gain and loss of electrons for each reactant. In this case the lowest common factor is 30. So the ClO3- must be multiplied by 5 and the I2 must be multiplied by 3. We now have the reactants balanced as follows: Step 4: Balance the products using the coefficients from the reactants. (Don’t change the coefficients for the reactants since they are already balanced!!) 000_Electric Cells Student Notes SS.doc 7 of 20 Step 5: Balance the oxygen atoms by adding H2O(l) (since the reaction happens in an acidic aqueous solution) Step 6: Since the reaction occurs in an acidic solution, H+(aq) ions are added to finish balancing the redox reaction. Type 3: Redox Reaction in an basic solution Methanol reacts with permanganate ions in a basic solution. Balance the redox equation for this reaction. Write the equation using the known reactants and products. Step 1: Assign each atom its oxidation number and identify the ones that change. Step 2: Determine the change in electrons per molecule involved. CH3OH(aq) has only one carbon atom, MnO4-(aq) has only one manganese atom, Step 3: Find the lowest common factor to balance the gain and loss of electrons for each reactant. In this case the lowest common multiple is 6. So the CH3OH(aq) is not changed at all and the MnO4-(aq) must be multiplied by 6. We now have the reactants balanced as follows: 000_Electric Cells Student Notes SS.doc 8 of 20 Step 4: Balance the products using the coefficients from the reactants. (Don’t change the coefficients for the reactants since they are already balanced!!) Step 5: Balance the oxygen atoms by adding H2O(l) (since the reaction happens in an basic aqueos solution) Step 6: Balance the hydrogens by adding H+(aq) ions. Step 7: Because this reaction happens in a basic solution, add an equal number of hydroxide ions (OH-(aq)) to both sides of the equation. Step 8: Combine any H+(aq) ions and OH-(aq) ions on the same side of the equation to make H2O(l) and cancel any ‘excess’ water. Method 2: Balancing Redox Reactions using Half-Reactions This method is very similar to that of balancing redox reactions using oxidation numbers. In this case you balance each part of the reaction separately (i.e. balance the reduction part by itself and then the oxidation part by itself). Example: Dichromate ions react with an acidic solution of iron(II) ions. The products formed are iron(III) and chromium(III) ions as shown by the following skeleton equation. Fe2+(aq) + Cr2O72-(aq) Fe3+(aq) + Cr3+(aq) Step 1: Separate the equation into its two component parts (a reduction and an oxidation). 000_Electric Cells Student Notes SS.doc 9 of 20 Step 2: Balance all the atoms other than oxygen and hydrogen in each halfreaction equation. Step 3: Balance the oxygen atoms in each half-reaction by adding H2O(l) and then balance the hydrogen atoms by adding H+(aq). Step 4: Balance each half-reaction separately by adding electrons to balance the charge. Step 5: Balance the two equations to each other by multiplying by a common factor to balance the number of electrons in each equation. Step 6: Add the two half-reactions together to get the final balanced equation. Cancel the electrons on both sides and any other common items on each side of the equation. For basic solutions: Step 7: For reactions that occur in basic solutions, simply add OH-(aq) to both sides of the equation to equal to the number of H+(aq) ions added. Step 8: Combine the H+(aq) and OH-(aq) ions on the same side of the equation to create H2O(l) and cancel any excess H2O(l) on either side of the equation. Homework: Pg. 668 #1, 2, 3, 4 and Pg. 673 #6, 7, 8 000_Electric Cells Student Notes SS.doc 10 of 20 Predicting Redox Reactions Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g) In this reaction the Fe2O3(s) is ________to Fe(s). The CO(g) used in this reaction is called the_____________. Reducing agent (RA)– 2Al(s) + 3Cl2(g) 2AlCl3(s) In this reaction the Al(s) is _________to AlCl3(s). The Cl2(g) used in this reaction is called the ____________. Oxidizing agent (OA)– Through a vast number of experiments a redox table (see Appendix C11) has been constructed to outline the relative strengths of reducing agents and oxidizing agents. Using this table reliable predictions can be made regarding whether or not a redox reaction will occur when these two substances are mixed together. Organization of Redox Tables Redox Spontaneity Rule 000_Electric Cells Student Notes SS.doc 11 of 20 From the table in Appendix C11, fluorine gas (F2(g)) is the strongest oxidizing agent (it is the least likely to give up electrons) and conversely, this makes it the weakest reducing agent (the most likely to gain electrons). Li+(aq) is the weakest oxidizing agent (it is the most likely to give up electrons) and conversely, this makes it the strongest reducing agent (the least likely to gain electrons). In general, Homework: Page 679 #14, 16, 18, 19 Predicting Redox Reactions in Solution Step 1: List all of the possible entities that are present in the solution. Step 2: Label all of the possible entities as either oxidizing or reducing agents. Step 3: Use this information and the spontaneity rule to determine if the reaction is spontaneous. 000_Electric Cells Student Notes SS.doc 12 of 20 Example: A solution of potassium permanganate is poured into an acidified iron(II) sulfate solution. Will there be a spontaneous redox reaction? Step 1: List all of the possible entities that are present in the solution. Step 2: Label all of the possible entities as either oxidizing or reducing agents and identify the strongest oxidizing agent and the strongest reducing agent. Step 3: Use this information and the spontaneity rule to determine if the reaction is spontaneous. Homework: Page 683 #1 6, 11 000_Electric Cells Student Notes SS.doc 13 of 20 Technology of Cells and Batteries When electrons move from one area to another they produce an electric current (electricity!). Since redox reactions involve the transfer of electrons, these chemical reactions can be used to produce electricity. Electric Cell – Battery – The positive electrode is called the The negative electrode is called the The amount of electrical energy that a cell can produce is dependent on the Voltage is a ratio of energy to charge. It is not dependent on the size of the cell but rather the chemical composition of the reactants in the cell. The total charge transferred by the movement of electrons in a cell/battery is measured in coulombs (C). Electric current (amperes (A) or coulombs/second) The larger the cell, the greater the current that can be produced by the cell. Example: The are many sizes of batteries (A, AA, AAA, C, D). All of these are 1.5V batteries. However, the bigger batteries can produce more current. Why do batteries have a ‘best before’ date? 000_Electric Cells Student Notes SS.doc 14 of 20 Inside the battery itself, a chemical reaction produces the electrons. Electrons flow from the battery into a wire, and must travel from the negative to the positive terminal (anode to cathode) for the chemical reaction to take place. That is why a battery can sit on a shelf for a year and still have plenty of power -- unless electrons are flowing from the negative to the positive terminal, the chemical reaction does not take place. Once you connect a wire, the reaction starts. The power (watts (W)) of a cell or battery The energy density or specific energy of a battery is This has become an important measurement for technological reasons. For example if the energy density is low (ie. lots of joules but a heavy mass as well) the battery would not be very good for a car. If the energy density is high (i.e. lots of joules but a low mass) if would be much more practical to be used in a car. Batteries that are made up of cells that can be used only once are called Batteries that are made up of cells that can be reused (rechargeable) are called Fuel cells are a special type of primary cell, Homework: Read pages 691 – 693 and make a table with the following headings: Fuel cell name Reactant at cathode Reactant at anode Electrolyte Potential uses Advantages Disadvantages 000_Electric Cells Student Notes SS.doc 15 of 20 Galvanic Cells Galvanic Cell – A half-cell consists of There are two basic designs of galvanic cells. One uses a ‘salt bridge’ which is a Ushaped tube filled with a non-reactive electrolyte (ie. positive and negative ions that are not part of the net reaction). Cotton plugs help to prevent the mixture of the two reacting solutions, but allow ions to pass. The other uses a porous porcelain cup that contains one of the electrodes and electrolytes. This is immersed in a beaker that has the other electrode and its corresponding electrolyte. (see page 695 in the textbook) Galvanic cells can be represented using a shorthand cell notation: The single line (|) indicates a phase boundary (between the electrode and its electrolyte) and the double line (||) indicates a physical boundary (the salt bridge or a porous porcelain cup). Example 1: For the following galvanic cell, determine which metal is the cathode and which metal is the anode. Cu(s) | Cu(NO3)2(aq) || Zn(NO3)2(aq) | Zn(s) 000_Electric Cells Student Notes SS.doc 16 of 20 Example 2: For the following galvanic cell, determine which metal is the cathode and which metal is the anode. Cu(s) | Cu(NO3)2(aq) || AgNO3(aq) | Ag(s) Explaining Observations in Galvanic Cells Evidence and Interpretation of the Copper-Zinc Cell Evidence Interpretation Copper electrode increases in mass; intensity of the blue colour of the electrolyte decreases The zinc electrode decreases in mass A voltmeter indicates that the copper electrode is the cathode (positive) and the zinc electrode is the anode (negative) The salt bridge (or the porous porcelain cup) is important to maintain electrical neutrality in each half-cell. This allows ions to pass to compensate for the electrons being lost from the anode or being gained by the cathode. Without this the galvanic cell would stop after a very short period of time. Galvanic cells with inert electrodes ________ and __________are commonly used as inert electrodes. These inert electrodes simply provide a surface for the half-reaction to occur. Homework: Page 700 #3,5 000_Electric Cells Student Notes SS.doc 17 of 20 Standard Cells and Cell Potentials Standard Cell – Standard Reduction Potential (Ero) – Standard Cell Potential (ΔEo) – Reference half-cell – Using Appendix C11 (Relative Strengths of Oxidizing and Reducing Agents), the standard cell potential for any two half-cells can be determined. In addition, the cathode and anode can be identified as well as the direction of the spontaneity of the reaction. Example 1: For the following galvanic cell, determine the identity of the cathode and anode, calculate the standard cell potential and write the net equation for the reaction. C(s) | Cr2O72-(aq) H+(aq) Cr3+(aq) || Pb2+(aq) | Pb(s) Step 1: find the half-cell reactions in Appendix 11 Cr2O72-(aq) + 14H+(aq) + 6e- Pb2+(aq) + 2e- Pb(s) 000_Electric Cells Student Notes SS.doc Cr3+(aq) + 7H2O(l) Ero = 1.23 Ero = -0.13 18 of 20 Step 2: Determine the identity of the cathode and anode. Since the chromate ion reaction has the higher reduction potential, it is the ________ in this reaction. Since the lead reaction has a lower reduction potential, it is the ______ in this reaction. NOTE: Step 3: balance the equations by finding the lowest common multiple for the number of electrons transferred (in this case 6) Example 2: A voltmeter indicates that in a standard copper-scandium galvanic cell (using Cu2+), copper is the cathode and the standard cell potential is 2.36V. Write the half-reactions for the two cells, the net equation, the shorthand cell notation for the cell, and determine the standard reduction potential for the scandium ion (Sc3+). 000_Electric Cells Student Notes SS.doc 19 of 20 Step 1: Write the half-cell reactions for the copper and scandium. Step 2: Find the lowest common multiple to balance the electrons in the two equations (in this case 6). Multiply the copper half-reaction by 3 (do not change the standard reduction potential value). Multiply the scandium half-reaction by 2. Add the two reactions together for the net reaction. The shorthand cell notation for this galvanic cell is: The standard reduction potential for the scandium ion (Sc3+) is: Homework: Pg 708 # 000_Electric Cells Student Notes SS.doc 10,11,12,14 20 of 20