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Ch.20 - Electrochemistry

Chapter 20, Problem 59a

Determine whether or not each metal dissolves in 1 M HNO3. For those metals that do dissolve, write a balanced redox reaction showing what happens when the metal dissolves. a. Cu

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everyone in this example, we're asked whether silver medal will dissolve in one Moeller nitric acid if yes, right. The bounce redox reaction that occurs upon dissolving this metal. So what we should recall is that we're going to need to refer to our standard electrode potential table in our textbooks or online. And we would see that nitric acid is going to oxidize as our nitrate an ion. And on our table we would see that this nitrate and ion goes through the following oxidation reactant reaction where we have formals of age plus reacting as well as three electrons added on which will produce our product nitrogen monoxide gas as well as two moles of water. This all comes from our standard electrode potential table and we would see that this has a value of 20.96 volts as it shows the reduction of our H plus proton. Now we need to consider silver on this table and we would see that silver, It's produced from our silver plus one caddy on gaining an electron here And on this table we have a value equal to 0.80V. And we would see that because 0.80V is less than 0.96V, we would say therefore silver will dissolve In one Moeller nitric acid. So this would be our first answer here because we've determined that because silver is lower on the standard electrode potential table in our textbooks and it has only a value of 0.80 volts than it will dissolve in the one molar nitric acid and now we need to write out our balanced redox reaction. So what we should see is that for our first half reaction we have three electrons on the reactant side. However, for our second half reaction here, we have just one electron. And so we want the electron numbers to match for both equations. So we're going to multiply this equation here by three. Sorry, this would be So multiplying by three here and so adding up these two equations, we would have our first equation added on to our second equation, which again is now going to be since we multiplied everything by 33 moles of silver producing three moles of the plus one silver cat ion Plus three electrons. And now our electrons are matching on both equations so we can cancel them out. And for our balanced redox reaction, we're gonna have our net equation, which is N 03 minus plus our formals of our H plus proton Plus three moles of our silver metal. And then on our product side we would now have our one mole of nitrogen monoxide gas plus two moles of water. And then from our second equation, we can now add on that three electrons that we bounced out. Or sorry, that would be the three moles of our A. G plus catalon. And so this here is going to be our second final answer for the balanced redox reaction when silver metal dissolves in one molar nitric acid. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below, and I will see everyone in the next practice video.