Can someone check this for me please?

[rocomath]

okay is this one right?
2Ag + 2e- -> 2Ag0

isnt the cathode reaction: Ag+ + e- -> Ag
and the anode reaction: Ag-> Ag+ + e-

Is that significant at all?

correct and yes b/c it tells you what role they're playing, but you're able to recognize them so it's all good.

anyways

so we have

$$Ag^+ (aq) + e^- \rightarrow Ag (s) \ E^0 = +0.80$$

$$Ni^{2+} (aq) + 2e^- \rightarrow Ni (s) \ E^0 = -0.28$$

which becomes

$$2Ag^+ (aq) + 2e^- \rightarrow 2Ag (s) \ E^0 = +0.80$$

DO NOT! multiply your charge by 2, Electrical potential is an intensive property just like density.

$$Ni^{2+} (aq) + 2e^- \rightarrow Ni (s) \ E^0 = -0.28$$

choosing Nickel to be our anode, we flip it's half-reaction

$$2Ag^+ (aq) + 2e^- \rightarrow Ag (s) \ E^0 = +0.80$$

$$Ni (s) \rightarrow Ni^{2+} (aq) + 2e^- \ E^0 = -0.28$$

notice how i didn't flip it's charge? there are two methods in dealing with whether you change your sign or not.

if you're using the equation $$E_{cell}=E_{cathode}-E_{anode}$$ keep your charges the same, if you're just going to add it all up without the equation, flip your charge with respects to the half-reaction you have chosen to be your anode.

so how would you write your overall reaction? (just add them up and cancel the electrons) and what is the EMF?

 

[yoshi6]

okay, i see...

Ag+(aq) + 2e- -> Ag(s) E0 +0.80V
Ni(s) -> NI2+(aq) + 2e- E0 -0.28V
_______________________________________

2Ag + Ni -> Ag + Ni2+ E0 0.52V

 

[yoshi6]

Like so?

 

[rocomath]

okay, i see...

Ag+(aq) + 2e- -> Ag(s) E0 +0.80V
Ni(s) -> NI2+(aq) + 2e- E0 -0.28V
_______________________________________

2Ag + Ni -> Ag + Ni2+ E0 0.52V

no, check your work again.

show me how you plugged your E0 into your equation or if you just flipped it.

 

[yoshi6]

did i add them wrong?

 

[rocomath]

did i add them wrong?

yep

show me your work if you were to do it with the equation

 

[yoshi6]

Ecell = E cathode - E anode
= .80 - (-0.28)
=1.08 ?

 

[rocomath]

Ecell = E cathode - E anode
= .80 - (-0.28)
=1.08 ?

correct

and if you just flip the equation and choose not to use the equation, you would've added and resulting in the same answer = +1.08

 

[yoshi6]

okay, awesome...I think I have some studying to do before first year starts...

 

[rocomath]

okay, awesome...I think I have some studying to do before first year starts...

alright, good luck! have fun

 

[rootX]

okay, awesome...I think I have some studying to do before first year starts...

chemistry
I also need to do some chemistry ; I have been planning to do it since last two months.
What are you going into, and if you don't mind, where?

If you are planning to really do it; then, you may try
http://www2.ucdsb.on.ca/tiss/stretton/CHEM2/organicx.htm

It has notes for all grade 12 topics/units!

 

[yoshi6]

Actually I only have to do it in first year. I am majoring in wildlife biology at the university of guelph.

 

[rocomath]

Actually I only have to do it in first year. I am majoring in wildlife biology at the university of guelph.

wow! beautiful campus ...

 

[yoshi6]

you've been? Lol don't you live in texas?

 

[rocomath]

you've been? Lol don't you live in texas?

i can be anywhere anytime ... <google>

it's nice tho, how's the weather? lol, it's way too hot down here.

 

[yoshi6]

ah isee...it is a really nice campus...well, I seem to think it is really hot right now, it's about 31 celsius, what about over there?

 

[yoshi6]

okay I have one last question that I am having trouble with. I am almost done my assignment. This is the last one...I have some examples in my book I am trying to follow, however, this is a little different.

If the Ka= 1.8 x 10^-5 for acetic acid (i.e . CH3COOH), what is the H+ ion concentration in a solution of this acid, if 1.2 grams of acid are dissolved in 1.0 L of solution.

So I assume I have to do this:

Ka= [H+][A-]/ [HA] ? and from there...

 

[rocomath]

so you have 1.2 grams of acetic acid, and you're told it's dissolved in 1.0L sol'n, what would your next step be?

your Ka expression is correct, question tho, is H+ and H3O+ the same?

by analyzing the Ka, what would you estimate the pH to be?

 

[yoshi6]

yes it is, isn't it?

 

[rocomath]

yes it is, isn't it?

yes, what would you estimate your pH to be?

$$CH_3COOH (aq) + H_2O (l) \leftrightarrow H_3O^+ (aq) + CH_3COO^- (aq)$$

what values you would input in your EQUIL for your ICE table?

 

[yoshi6]

okay

Ka = 1.8 x 10^-5= (9.0 x 10^-3)^2
_____________
[CH3COOH]

At = , [CH3COO-] = [H3O+] = 9.0x10-3mol/L

I think...

 

[yoshi6]

okay that is supposed to be (9.0 x 10^-3)^2 divided by [CH3COOH]

 

[rocomath]

okay

Ka = 1.8 x 10^-5= (9.0 x 10^-3)^2
_____________
[CH3COOH]

At = , [CH3COO-] = [H3O+] = 9.0x10-3mol/L

I think...

where did you get 9.0 x 10^-3?

 

[yoshi6]

wrong? I don't know I'm very confused, I am trying to follow the examples in my book, but I don't really understand what it going on...

 

[rocomath]

$$K_a=\frac{[H_3O^+][CH_3COO^-]}{CH_3COOH}$$

$$1.8\times10^{-5}=\frac{x^2}{\frac{[1.2/60.06]}{1.0}}$$

you wrote your Ka correctly up there, idk what happened.

you're correct that the Acetate ion and Hydronium ion concentrations will be the same.

 

[yoshi6]

but where does the 60.06 come from? what is that

 

[rocomath]

but where does the 60.06 come from? what is that

let me fix that, i should have that within another fraction

how do you calculate molarity? what units should/are used for equilibrium concentrations?

 

[yoshi6]

molarity = moles of solute/ liter of solution

 

[rocomath]

molarity = moles of solute/ liter of solution

what is the molar mass of Acetic acid and the moles of 1.2 grams of Acetic acid that will be dissolved?

 

[yoshi6]

60.05 g/mol i think