Question

Oxidation Half-Reaction: Zn(s) —> Zn2+ + 2e-
Reduction Half-Reaction: NO3-(aq) + e —> NO2(g)
      iv. Equalize charges for the two half-reactions. (1 point)
      v. Add the equations and simplify to get a balanced equation. (1 point)

Answer 1

@SolomonZelman

Answer 2

I really don't know. I am not good at chemistry-:( I used to be good at balancing these equations, but....

Answer 3

I am sorry-:( .. seriously (!)

Answer 4

its okay thanks anyways

Answer 5

This is pretty simple,

LEO the lion says, GER

lose electrons oxidation
Gain electrons Reduction

You should have an equal gain and lose of electrons on both sides of the equation

Answer 6

In this reaction something is gaining electrons and something is losing electrons. Half reactions just show what is gaining electrons and what is losing electrons.

Answer 7

Do you know how to solve this now?

Answer 8

im not sure how to balance the charges

Answer 9

When you are adding two reactions you simply have to cross out what is on both sides of the equation for example,

Half Reaction:

Zn(s) → Zn2+ + 2e−

Cu2+ + 2e− → Cu(s)

Added together

Zn(s) + Cu2+ + 2e− → Zn2+ + 2e− + Cu(s)

notice that 2e- is on both sides of the equation, thus they can be classified as redundant, same goes with reactants and products if they are present on both sides of a chemical equation when you add the half reactions they cancel out (they are redundant) and you dont have to and shouldn't write them.

Final Reaction:

Cu2 + Zn(s) → Cu(s)


Notice that the electrons are on both sides of the equation

Answer 10

the final line was a mistake in editing ignore it.

Answer 11

okay thanks

Answer 12

You should notice that the charges are not equal on the half reactions, one reaction is giving 2 electrons the other is only giving one you can either divide one by 2 or multiply another by 2 to equalize their charge

Answer 13

wait so how can i fix it to make the charges equal for the oxidation and reduction

Answer 14

https://en.wikipedia.org/wiki/Half-reaction

Answer 15

I just told you how to fix the charges so they are equal

Answer 16

oh okay i see it

Answer 17

so to equalize them it would be:
2 Zn(s) —>2 Zn2+ + 4e-
NO3-(aq) + e —> NO2(g)

Answer 18

i multiplied the top one by two.

Answer 19

how does that makes sense

Answer 20

when you add the two reactions together you should not have any electrons in the equation

Answer 21

If you do they are not equalized in charge

Answer 22

i have no idea how to do this...

Answer 23

so i have another question...
c. If 852.04 g Al(NO3)3 reacted with the 741.93 g Na2CO3, which would be the limiting reagent? (2 points)
d. Use the limiting reagent to determine how many grams of Al2(CO3)3 should precipitate out in the reaction. (2 points)
e. If only 341.63 g of Al2(CO3)3 precipitate were actually collected from the reaction, what would the percent yield be? (2 points)

Answer 24

2 Al(NO3)3 + 3 Na2CO3 = Al2(CO3)3(s) + 6 NaNO3

Answer 25

https://www.youtube.com/watch?v=vDdVWUVe-pY

Answer 26

which one is correct? I tried solving it two times and I'm not sure.

Answer 27

https://www.youtube.com/results?search_query=half+reactions

Answer 28

can you check my screenshots? thanks

Answer 29

The limiting reagent is simply the reactant that runs out first, so all I do to find it is either use logic or I just use the balanced equation to figure out which will give me the least amount product

Answer 30

The one that gives you the least product is your limiting reagent

Answer 31

no but i have to use the limiting reagent to find the percent yield and stuff. however, in the first on i got 2 and in the second one i got 4.33

Answer 32

is the first screenshot or the second one right..? i know one of them has to be

Answer 33

According to your calculation you have,
7 moles of Na2CO3
4 moles of Al(NO3)3

The reaction is:
2 Al(NO3)3 + 3 Na2CO3 = Al2(CO3)3(s) + 6 NaNO3

So,

Using Al(NO3)3:
(4moles/2)*1 = 2moles of Al2(CO3)3(s)

Using Na2CO3
(7moles/3)*1 = 2.33 moles of Al2(CO3)3(s)

Therefore Al(NO3)3 is the limiting reagent

Answer 34

you can do the same thing with NaNO3 and you would get the same result

Answer 35

wait so is the percent yield supposed to be 31 or 73

Answer 36

Percent Yield = (Actual Yield/Theoretical Yield)*100

Answer 37

You can either use grams or moles in that equation doesnt matter

Answer 38

oh okay that makes sense. thanks

Answer 39

i got it

Answer 40

Easy right?

Answer 41

Do you understand everything? If you have any questions ask now

Answer 42

or anymore questions

Answer 43

yep

Answer 44

What part of an "electrochemical cell" is the baking soda, salt, and water mixture? i know theres the salt bridge and electrolyte and stuff

Answer 45

https://upload.wikimedia.org/wikipedia/commons/2/2f/Galvanic_cell_with_no_cation_flow.png

Look at this, and look at LEO the lion says GER

Answer 46

You should be able to figure out what the anode and cathode is, but if you have not learned about electromotive force you can probably just call it a reactant

Answer 47

or you could actually just write out the reaction of sodium bicarbonate and water