Can someone check this for me please?

[yoshi6]

Can someone check this for me please??

The solubility of silver sulphate is 1.3 x 10^-6 mol/L at 20 degrees celcius.

a) write the equation for the dissolving of this salt in water.
b) show the Ksp expression.
c) calculate the value of the Ksp at 20 degrees celcius.

ANSWERS

a) 2Ag+(aq) + SO4 2-(aq) = Ag2(SO4)

b) Ksp = [Ag+]^2[SO4^2-]

c) I am having trouble withthis: Ksp= (1.3 x 10^-6)(1.3 x 10^-6) = 1.7 x 10^-12

 

[rocomath]

lol wrong place. let me work it out.

 

[yoshi6]

lol thanks...and I'm a girl! :)

 

[rocomath]

lol thanks...and I'm a girl! :)

lol i thought about that when i said bro, so i edited my comment but you checked it fasttt! haha

anyways, so I'm assuming C is your work?

$$Ag_2SO_4 (s) \rightarrow 2Ag^{+} (aq) + SO_4^{2-} (aq)$$

You're given $$K_{sp}=1.3x10^{-6} M$$ of Silver Sulfate, and so when you write your $$K_{sp}$$ equation it will look like this. (You have it correct, I'm just blabbering)

$$K_{sp}=[Ag^{+}]^2[SO_4^{2-}]$$ and $$Ag_2SO_4 (s)$$ is not included b/c we don't include liquids/solids since their concentration change is negligible.

Let [tex]x=1.3x10^{-6}, x=[2S]^2[/tex]

simplifying and skipping steps

$$1.3x10^{-6}=4S^{3}$$

$$S=\sqrt[3]{\frac{1.3x10^{-6}}{4}}$$

and that's all.

 

[yoshi6]

Ok. Well first, I have this one question I have posted before. I really have no clue what they are asking me, I do not know what to do or where to start. Maybe if you have any ideas I can work with it...

A solution contains the following ions, each at a 0.1 M concentration.

Ag+ (aq), Ca2+ (aq), Ni2+ (aq)

write out a procedure by which these ions may be separated from each other and from the solution, clearly indicating the order of separation and when filtering should be done.

So, what is this procedure and order of separation they are talking about? I have no clue what is going on here.

 

[rocomath]

i'm not very good at these type of questions (i didn't care for labs too much, I'm lazy), but what comes to mind is solubility rules & activity series of metals.

i'm sure someone will respond :-]

 

[yoshi6]

OK, and I am having the same kind of issues with this one:

An electrochemical cell is constructed by placing a nickel electrode into a 1.0 M NiSO4 solution, and a silver electrode into a 1.0 M AgNO3 solution, and then joinging them with a salt bridge to complete the circuit.

a) give the anode half reaction and the E0 value.
b) give the cathode half reaction and the E0 value
c) give the net reaction and its E0 value

 

[rocomath]

i think i have an idea

$$Cl^{-}$$ is soluble except in the presence of $$Ag^{+}$$ in which it is a solid.

so add $$Cl^{-}$$ to react with Ag, so now you've taken care of Ag -> AgCl (s)

 

[rocomath]

OK, and I am having the same kind of issues with this one:

An electrochemical cell is constructed by placing a nickel electrode into a 1.0 M NiSO4 solution, and a silver electrode into a 1.0 M AgNO3 solution, and then joinging them with a salt bridge to complete the circuit.

a) give the anode half reaction and the E0 value.
b) give the cathode half reaction and the E0 value
c) give the net reaction and its E0 value

read this thread real fast https://www.physicsforums.com/showthread.php?t=181411

and i will work on your problems while thinking of some questions to ask

 

[rocomath]

ok i got it worked out, my question is. what is the purpose of Sulfate and Nitrate?

what are the charges for the individual ions? Ni, SO4, Ag, NO3?

 

[yoshi6]

The common charges?

Ni2+, Ag1+, SO4 2-, NO3 1-...right?

 

[rocomath]

The common charges?

Ni2+, Ag1+, SO4 2-, NO3 1-...right?

EDIT: lol ... you have it right, idk what I'm thinking, anyways moving on

so did you read the other thread?

so now look at the reduction potential table and tell me which one you would choose to be your cathode and anode?

 

[yoshi6]

lol btw the four of them?

 

[yoshi6]

or just between NiSO4 and AgNO3

 

[yoshi6]

I would have to say AgNO3 cathode...

 

[rocomath]

you didn't answer my question, what is the purpose of Nitrate and Sulfate?

look at your reduction potential table, are you able to find them?

why do you choose AgNO3 as your cathode?

 

[yoshi6]

I am so confused

 

[rocomath]

I am so confused

sigh :-[ ok let's move on

a rule of thumb, $$E_{cell}>0$$, we need to choose which one will be the cathode/anode so that $$E_{cell}$$ is positive.

tell me the what the half-reactions for $$Ag^{+}$$ and $$Ni^{2+}$$.

 

[yoshi6]

Okay, like so?

Ag -> 2g-
is that right for Ag+

 

[yoshi6]

Ag-> Ag+ + e- is oxidation right?

 

[rocomath]

Okay, like so?

Ag -> 2g-
is that right for Ag+

ok from this list, which half-reactions for Ag+ and Ni2+ would you choose?

 

[rocomath]

Ag-> Ag+ + e- is oxidation right?

yes, correct.

 

[rocomath]

ok so our half-reactions are

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

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

correct?

 

[yoshi6]

Okay so
Ni2+(aq) + 2e- ---> Ni(s) - .28

Ag+(aq) + e- ------> Ag(s) .799

 

[yoshi6]

lol wow, I'm embarressed

 

[rocomath]

Okay so
Ni2+(aq) + 2e- ---> Ni(s) - .28

Ag+(aq) + e- ------> Ag(s) .799

now balance your equation so that the amount of electrons being transferred is the same.

then choose which will be the anode b/c they are currently both in their reduction state.

$$E_{cell} > 0$$

 

[rocomath]

lol wow, I'm embarressed

don't be :-] i don't know anything about labs so I'm useless there :-D

 

[yoshi6]

sorry my computer keeps shutting down...one minute

 

[rocomath]

btw, i think most tables are ordered from the most positive value (reduction, oxidizing agent) to the most negative (oxidized, reducing agent)

so be careful

 

[yoshi6]

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?

 

[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?