Balance each redox reaction occurring in acidic aqueous solution....

Question

Balance each redox reaction occurring in acidic aqueous solution.
a. K(s) + Cr3+(aq) ¡ Cr(s) + K+(aq)

Accepted Answer

hey everyone in this example, we need to balance the given redox reaction under acidic conditions. So what we should recognize is that because this is a redox reaction, we need to have two half reactions. So our first half reaction will come from our first reactant being the solid sodium which is going to produce our product where we form the one mole of the plus one Catalan of sodium. So we should recognize that we went from a net neutral charge on the reactive side to a net plus one charge on the product side, meaning that we lost electrons. And so this would be the oxidation half of our reaction. Moving on to the second half reaction, which comes from our second reactant. We have three moles of the iron, three plus candy on or sorry, one mole of the iron three plus Catalan which produces one mole of solid iron. And so sorry this says As for solid. And so what we can see is that we went from a net three plus charge on the reactant side to a net neutral charge on the product side, meaning we gained electrons. And so this would be the reduction half of our given reaction. So now our next step is to make sure that our atoms are balanced for both of our half reactions, we have one mole of the sodium atom on the reactive side and one mole on the product side. So the sodium is balanced in the first half reaction. And we have the same case going on in the second half reaction and we have one mole of each atom. So because all of our atoms are balanced, we can now focus on balancing out the net charges here. So looking at our first half reaction, we have a net charge of zero on the reactive side and a net charge of plus one on the product side, meaning we need to get rid of this plus one charge on the product side here. And so we're going to get rid of that plus one charge by gaining an electron. So we would say plus one electron. Now for the second half reaction we have a net charge of plus three on the reactant side and a net charge of zero for our neutral iron on the product side, meaning we need to cancel out this three plus charge. So we're going to cancel out this plus recharge by expanding our react inside here. So that we say we can gain three electrons. So now that we have done that, we need to make sure that these electrons in both of our half reactions are now balanced or rather equal to one another. So we need to go ahead and multiply this first half reaction by a factor of three. And that would give us three electrons so that the electrons match. And so what we can say we now have is three moles of solid sodium producing three moles of the sodium plus one catalon plus three electrons. And then for our second half reaction, we would still maintain the same reaction where we have one mole of the three plus iron carry on Plus three electrons producing our one mole of solid iron. And so now we can go ahead and cancel out what appears on both or the opposite side of both equations, which would be our number of electrons where we have three electrons on the product side for the first half reaction, and three electrons on the reactive side for the second half reaction. And now we would add these two reactions up now that we've canceled out the electrons so that we now have our balanced net redox reaction, which is three moles of the solid sodium atom Added to one mole of the iron, three plus catalon producing one mole of our solid iron, Plus three moles of our sodium catalon. And so this would complete this example here as our final answer for our balanced redox reaction under acidic conditions. So I hope that everything I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video

Balance each redox reaction occurring in acidic aqueous solution. a. K(s) + Cr3+(aq) ¡ Cr(s) + K+(aq)

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