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Balancing Redox Reactions Practice Worksheet Key Problem #1

Added Oct 02, 2014 •

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Balancing a redox reaction under acidic conditions.

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okay this is mr. Mata for AP Chemistry it is October 1st 2014 and this is the answer key for can you balancing redox reactions practice worksheet so this is question 1 again this is acidic conditions that were looking at here so our first step normally be to actually take the redox reaction and to break it up into two half reactions so let's go ahead and take the manganese first the manganese CentraCare we got the permanganate ion now let's go ahead and do the electron change here well let's go into the other part and we've got the bismuth air and  sorry acting side and then we have the business ion the product side  Snell is going to see what's going on with the electrons here those are changing so basically we've got  the manganese Ion at + 2 and it's going to change on this Riverside here let's see how it changes now again I member these have to all add up the oxidation states to -1 because the permit at ion is a negative charge  so there is -1 so we know oxygen is normally -2 so right there we got -2 and 4 that's negative 8 and that means the manganese would have to be plus 7 so we have those together we get an overall charge of -1 which is the ion  for the Mega. Church  chug wine from a  + 2 to a + 7 that is losing five electrons or again oxidations going on there and so what we're going to do is we're going to go ahead since they're being lost oxidation over here  cancer going to actually put our five electrons over here Humvee  products oxidation  now for going to look at the business half-reaction it's a product at + 3 summer thing over here and again the overall charge is going to be minus one we got a minus two for the oxygen against that's -6 and so the Christmas must be a plus 5 so overall we get a negative one charge  Subway in from + 5 - + 3 that would be a gain of two electrons  holy say reduction  silver head and make the reduction I'm at Park down there and again will add the two electrons on the  reactant side now we're going to go in and take care of the oxygen so you know it's the both half reactions have oxygens here in the for the main Ion on the product side on the business side it's on the reactants is we have to make sure we can of those and of these to the conditions so before oxygen Siri going to counter those with for water put those over here on the left and that takes care of our for Oxygen's on both sides but we also increased our hydrogens by 8 on the left sore going to add 8 hydrogen ions on the right counter those and somewhere where I can do that for the business half we have three oxygens here on the reactant side so we're going to add three  water molecule can I get on that also means we have six hydrogen ions with a duster going to add those six hydrogen ions over here  on the reactant side and that takes care of the hydrogen and the oxygen now we're going to go ahead and make sure are electrons being oxidized reduced are the same and so to do that we looking we find here is a five electron and the oxidation and two for the reactant so we're going to find the least common multiple of 5 and 2 which is 10 and we're going to multiply the half reactions by a whole number to give us that value so for the top manganese 1 * 2 and for the bottom bismuth 1 x 5 and so we do that now we're going to spacely multiply through the little tiny bit go okay so take care of all the reactants first we've got to * 4 is 8 water molecules we have to manganese ions we have 5 * 6 is 30  hydrogen ions  5 * 2 is 10 electrons and then we have five times the business is 5 ions hand in for the Sonic side we go back to the top 2  permanganate Islands 10 electron and 16  hydrogen ions and I were going to take care of the bottom Rex over here five times the bismuth ions and then five times the Three Water's gives us 15 Waters so now there's going to be a c reduce and by reducing to take the waters first day of 8 on the reactant side 15 on the product side so we'll lose 8 water is there this becomes 7  on the product side  are 10 electrons on each side cancel out not make sure they're balanced as well  and then that should be the final canceling out here with the camera little bit there we go a lot of stuff going on there  and so when I finish out my entire  await hydrogen stay until 2:30 over here 16 over here still 16 or gone and now this over here becomes 14 now that should be good tried everything else I've got to  manganese ions  + 14 hydrogen ions + 5 the Bismuth ions two remain Zion's  5 bismuth ions and last but not least water molecules so that will be my final cancer inside the rectangle yes a lot of steps just go through the system acted on that's how you get the answer again that was problem the first problem under static Solutions  snickle item pause here and I will put up the second acidic problem

Uploaded by Edward Mata.

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Balancing Redox Reactions Practice Worksheet Key Problem #1

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