Chem 222 Week 12 notes
Chem 222 Week 12 notes Chem 222
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This 3 page Class Notes was uploaded by Leslie Pike on Friday April 22, 2016. The Class Notes belongs to Chem 222 at Western Kentucky University taught by Darwin Dahl in Spring 2016. Since its upload, it has received 17 views. For similar materials see College Chemistry 2 in Chemistry at Western Kentucky University.
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Date Created: 04/22/16
Balancing a redox reaction (acidic aqueous solution): Cr2O 72-+ Fe 2+ Cr 3++ Fe 3+ Split it into two half-reactions. 2- 3+ Cr2O 7 Cr Fe2+ Fe 3+ You will have to balance both equations. Balancing the second one is easier than balancing the first one, because you will already know which side the electrons go on. Start with the easier reaction (in this case, the one with iron). First, mass balance. (Already done.) Then, charge balance. Charge balancing looks intimidating when you first try it, but actually it’s just simple arithmetic. The left side and the right side have to equal. You have +2 on the left and +3 on the right. To make the equation balanced, subtract 1 from 3. +2 = +3 -1 Ta-da, it’s that easy. You add one electron to the right side of the equation, and you’re done! Now, for the other half-reaction. Mass balance it first. (You are in aqueous solution so you can add all of the water molecules you need, since you have plenty of them floating around. You are in acidic solution so you can add all of the hydrogen ions you need, since you have plenty of them floating around.) + 2- 3+ 14H + Cr O 2 7 2Cr + 7H 2 Now, you need to charge balance it. This equation looks more intimidating than the previous one, but you can just break it down into smaller pieces. Hydrogen, oxygen, and water do not participate in the actual electron transfer—they can be ignored. This leaves us with chromium. The chromium in dichromate has a charge of +6, and the free chromium ion has a charge of +3. DO NOT NEGLECT THE FACT THAT THERE ARE TWO CHROMIUM ATOMS IN DICHROMATE, AND TWO FREE CHROMIUM IONS ON THE RIGHT SIDE OF THE EQUATION. Write the amount of charge on each side of the equation, and subtract where you need to subtract to make both sides equal (electrons) + 2(+6) = 2(+3) -6 +12 = +6 You have to add six electrons to the left. Check your work; one of the half-reactions should have electrons on the right, and the other should have them on the left. (This is why you do the easy one first, so you can check the harder one against the easy one.) The electrons will need to cancel. You have 6 electrons in the chromium equation, and 1 electron in the iron equation. Multiply the iron equation by 6 so that the electrons cancel, and add the two equations. 14H + Cr O 2+ 6 Fe 2+ 2Cr 3++ 7H O + 6Fe 3+ 2 7 2 This is how to balance a redox reaction in acidic aqueous solution. For a basic aqueous solution, do the same thing. Then, after you have the final equation, add + hydroxide to both sides to cancel the H . For the above equation, this would be 14 hydroxide. You would have 14 water on the left, 7 water on the right, and 14 hydroxide on the right. Waters cancel each other, and you have 7 water on the left and 14 hydroxide on the right. There are two types of electrochemical cells: galvanic (voltaic) or electrolytic. In a galvanic cell, the reaction makes electricity. In an electrolytic cell, electricity makes the reaction. Galvanic cells have a positive EMF, and electrolytic cells have a negative EMF. Oxidation occurs at the anode, and reduction at the cathode. (The word “oxidation” and the word “anode” both start with vowels, and the word “reduction” and the word “cathode” both start with consonants.) Anodes shrink and cathodes grow. Oxidation is the loss of electrons (metal turns to aqueous ions), and reduction is the gain of electrons (aqueous ions turn to solid metal). When writing the shorthand notation for a battery, you start at the anode and end at the cathode. (A comes before C in the alphabet, so you do the anode first.) For the traditional zinc-copper battery: 2+ 2+ Zn (s) (aq)(x M)||Cu (aq)y M)|Cu (s) Here, x and y represent the concentrations of the zinc and copper ions, respectively. The standard EMF of the cell is calculated as: E0 = E 0 + (-E 0 ) cell cathode anode The cathode reaction will be going in the forward direction as it is written in a table. The anode reaction will be going in the reverse direction as it is written in the table; hence, its EMF will have an opposite sign of whatever the table says (this is why it has a negative sign in front of it in the equation). Standard conditions are 298 K, 1 M concentration and 1 atm. To determine the EMF of a cell not at standard conditions: E = E – (.0592/n)*log(Q) Calculate Q for your reaction, calculate E cellrom the E values of your cathode and anode, and determine the number of free electrons involved in the reaction. Then, plug everything in and solve. To convert from free energy to EMF: ΔG=-nFE F is the Faraday constant, 96,485 Coulombs/mole of electrons. Lowercase n is the number of free electrons involved in the reaction, E is cell EMF.
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