Consider the unbalanced redox reaction: Cr2O72-(aq) + Cu(s) ¡ Cr3...

Question

Consider the unbalanced redox reaction: Cr2O72-(aq) + Cu(s) ¡ Cr3+(aq) + Cu2+(aq)
Balance the equation and determine the volume of a 0.850 M K2Cr2O7 solution required to completely react with 5.25 g of Cu.

Accepted Answer

hey everyone in this example, we need to give the balanced equation for the following redox reaction as well as answer how much 0.3 84 molars of sodium nitrate solution in milliliters is needed to thoroughly react with 4.23 g of chromium. So based on this given redox reaction, we're gonna make two half reactions are first half reaction is going to be our solid chromium, which forms the chromium three plus carry on as a product. And then for our second half reaction, we have our nitrate and ion which is going to form nitrogen, monoxide and gaseous form as a product. Our next step is to make sure that our atoms are balanced, we have one atom of chromium on each side of the equation. So our atoms for this first half reaction is balanced. However, for our second half reaction we have one nitrogen on the reactive side and one nitrogen on the product side, but three oxygen on the reactive side and only one on the product side. So we should recall that we can balance out oxygen by using water. So we would add two moles of water to the product side to give us a total of three oxygen's on the product side to bounce out with the reactant side. Now we've added or by adding these two moles of water, we now have four moles of hydrogen on the product side. So we're going to add our to our react inside four moles of hydrogen H plus to be able to balance out our hydrogen on both sides of our equation. And so now that all of our atoms are balanced in both of our half reactions, we need to make sure our net charges are balanced as well. So for our first half reaction we have a net charge of zero on the reactive side and the net charge of plus three on the product side. And so this would mean that we need to go ahead and add not to our reactive side actually, but to our product side, three electrons so that we can have a net charge of zero. Now for our second half reaction we have a net charge on the reactant side of four plus plus minus one, which would give us a net charge of plus three on the reacting side. Whereas on the product side we have just neutral molecules. So we have a net charge of zero. And so we would go ahead and get rid of that plus three charge on the reactant side by adding three electrons to the reactant side. So that now we have a net charge of zero on both sides of this second half reaction. So now that we have our atoms as well as our net charge is balanced in both of our half reactions, we're going to add these two reactions up, we're gonna cancel out the electrons. And for our net balanced reaction, what we should get is one mole of solid chromium plus one mole of nitrate as our an ion Plus four moles of hydride yields one mole of nitrogen monoxide gas plus two moles of water Plus one mole of our chromium catalon. And so this balanced reaction would be our first answer to complete this example. And so now we can move on to part two of this equation or sorry, of this prompt, which asks us for the volume and milliliters of sodium nitrate needed to react with 4.23 g of chromium. So we're going to begin with that massive chromium which is given to us as 4.23 g of chromium. And we're going to convert from grams of chromium, two moles of chromium. So we should recall from our periodic table that we have a molar mass of 52 g of chromium for one mole of chromium. Now we're able to cancel out grams and we're gonna move from moles of chromium, two moles of nitrate as an an eye on. And we're going to sorry, do so by looking at for from our balanced equation above the ratio between chromium to nitrate. And we would see that we have coefficients of one here for each of these reactant. So we have a 1 to 1 molar ratio. So we would say we have one mole of chromium for one mole of nitrate. Now we're able to cancel our moles of chromium and now we're going to convert from molds of nitrate, two moles of sodium nitrate. Where we would recognize that in sodium nitrate we just have one mole of sodium nitrate in this. one mole of sodium nitrate. Sorry. So one mole of nitrate in our one mole of sodium nitrate here and now we're able to cancel out moles of nitrate, meaning now we can move for a last conversion factor or rather the second to last conversion factor to go from moles of sodium nitrate. two leaders of sodium nitrate. And just to make more room because we still have one more conversion factor from the prompt. We're going to interpret the given polarity because we recall that morality can be interpreted in moles divided by leaders. So we would say that we have 0. moles of sodium nitrate for one liter of sodium nitrate. This allows us to cancel out moles of sodium nitrate and now we have our last conversion factor to convert from leaders to middle leaders. So we should recall that our prefix milli tells us that we have 1/10 of the third power milliliters for one leader. Now we're able to cancel out leaders and we're left with ml as our final unit, which is what we want according to the prompt and what we're going to get is a volume equal to 212 mil leaders as the volume of sodium nitrate. That is needed to thoroughly react with our chromium. So everything highlighted in yellow represents our final answers. I hope that everything I explained was clear. If you have any questions, please leave them down below. Otherwise, I will see everyone in the next practice video.

Consider the unbalanced redox reaction: Cr2O72-(aq) + Cu(s) ¡ Cr3+(aq) + Cu2+(aq) Balance the equation and determine the volume of a 0.850 M K2Cr2O7 solution required to completely react with 5.25 g of Cu.

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