CHEMISTRY COMMUNITY

Charlyn Ghoubrial 2I

I’m pretty sure you’ll be given the info you need on the final

Charlyn Ghoubrial 2I

I’m pretty sure you would use an ICE table. Find out the molarity of what you can and plug in the rest to the ice table

Charlyn Ghoubrial 2I

Dr Lavelle answered this question a few years back and said “you separate chemicals in different phases using a single vertical line and if you have multiple chemicals in the same phase (say 2 aqueous solutions) you separate them with a comma. Then, you separate cathode and anode by using a double ...

Charlyn Ghoubrial 2I

Some solution guides have contradicting ideas but Dr Lavelle mentioned that species in the same state are separated by commas so I would just trust him on that

Charlyn Ghoubrial 2I

For cell potentials, you only reverse the signs if you reverse the equation (without multiplying by coefficients) because it’s an intensive property that’s independent of stoichiometric coefficients. For enthalpy you and entropy, since they’re extensive properties, you reverse signs when reversing t...

Charlyn Ghoubrial 2I

I’m pretty sure you use Ecell= Ecell*+(0.592V/n) x pH
Your Ecell* is 0, n=2, pH=11

Charlyn Ghoubrial 2I

N is the number of electrons that you cross out from both sides of the reduction and oxidation half reactions after you balance them

Charlyn Ghoubrial 2I

The organic chemistry tutor breaks it down really well. Some people prefer khan academy so you could try both but i personally understand more from the organic chem tutor

Charlyn Ghoubrial 2I

H is going from either 0 (because it’s by itself) to +1 and viseversa so only 1 electron is being transferred

Charlyn Ghoubrial 2I

You would flip the half reaction from appendix 2B (flip products and reactants) and flip the sign as well

Charlyn Ghoubrial 2I

I think it’s because your product is already H2O and everything is balanced except O2 so you add it to the left to balance on the right. Otherwise you would unnecessarily keep adding H+

Charlyn Ghoubrial 2I

I think that’s a typo because ln goes from 3+ to 2+ and U goes from 4+ to 3+ so in both cases, one electron is being transferred but I’m not 100% sure

Charlyn Ghoubrial 2I

Pt is added to any side that does not have a solid conductor so whenever you only see (aq) or (g) or no solid conductor, add Pt

Charlyn Ghoubrial 2I

It just implies that none of your species are solids. They’re either (aq) or (g) if you’re talking about example d

Charlyn Ghoubrial 2I

So you have to use the equation given to figure out the 2 half reactions you would use. One half reaction equation is your oxidized one (anode) and the other half reaction is your reduced one (cathode). When writing out cell diagrams, your anode always goes on your left and your cathode goes on your...

Charlyn Ghoubrial 2I

Sorry this question is about 6M.3... Can someone briefly explain the process for 6M.3? I just want to make sure I have the right concepts down Use the equation Ecell= Ec - Ea ( Ecathode - Eanode) and then figure out the standard reduction potential of each anode and cathode and then plug them in th...

Charlyn Ghoubrial 2I

Angela Patel 2J wrote:Also, how would we know the sign of the internal energy?

If energy enters the system: +
If energy leaves the system: -
If work is done on the system: +
If work is done by the system: -

Charlyn Ghoubrial 2I

I’m not entirely sure but since Br goes from -1 to +5 we know that this reaction is being oxidized so we just assume that O3 to O2 is being reduced I guess. But I don’t think O3 should pair up with BrO3- because that would put O2 with Br- and i don’t think that we can work with a half reaction like ...

Charlyn Ghoubrial 2I

Acidic: split equation in 2 half reactions and balance them (adding H2O to balance Os and H+ to balance Hs) then you cancel out your balanced electrons and anything else that needs cancelling and write your balanced equations Basic: same steps as Acidic but then you add OH- on the side that has H+ t...

Charlyn Ghoubrial 2I

The most ones you need to know for this class are: Any element standing by itself (unattached to another element) is always 0 O2 (when attached) is almost always 2- (H2O2 is an exception: O2 is -1) H is always +1 And from there you can figure out the rest from looking at equations The organic chemis...

Charlyn Ghoubrial 2I

If I’m looking at the correct question, your half reaction for oxidation (at anode) should be Sn^(2+) —> Sn^(4+) +2e- and your half reaction for reduction (at cathode) should be Hg2Cl2 + 2e- —> 2Hg + 2Cl- So then your overall balanced equation after you cancel out the 2e- will be Hg2Cl2 + Sn ^(2+) —...

Charlyn Ghoubrial 2I

Dr Lavelle answered this question a few years back and said “you separate chemicals in different phases using a single vertical line and if you have multiple chemicals in the same phase (say 2 aqueous solutions) you separate them with a comma. Then, you separate cathode and anode by using a double l...

Charlyn Ghoubrial 2I

When the anode or cathode does not have a metal (all aqueous)

Charlyn Ghoubrial 2I

I would just look at the appendix. The half reactions would be
Reduction: Ag+ + e- —> Ag
Oxidation: Ag + Br- — AgBr + e-

Charlyn Ghoubrial 2I

It is Ecell

Charlyn Ghoubrial 2I

Your half reactions should be
Reduction: O2 + 2H2O + 4e- —> 4OH-
Oxidation: 2H2O —> O2 + 4H+ + 4e-
Cancel out the O2, 4e-
Balanced equation: 4H2O —> 4H+ + 4OH- which reduces to H2O —> H+ + OH-

Charlyn Ghoubrial 2I

I don’t think you have to find Ecell. Just find the half reactions and the balanced equation for each

Charlyn Ghoubrial 2I

A lot of people have also been asking similar questions and it seems like no one really knows so either Dr Lavelle will provide us with the information on the test or I’m just going to try to see if there is a pattern to which one to use from all the questions and answer key in the textbook

Charlyn Ghoubrial 2I

An easy way to do this problem is to use the equation Ecell=Er-El
Ecell= E(Cu^(2+)/Cu) - E(M^(2+)/M)
E(M^(2+)/M)= (-0.689+0.34)V = -0.349V

Charlyn Ghoubrial 2I

If delta G is negative, your reaction will be spontaneous so the metal will dissolve and if delta G is positive, your reaction isnt spontaneous so the metal will not dissolve

Charlyn Ghoubrial 2I

I’m actually not sure. My half reactions were
Reduction: Au+ + e- —> Au
Oxidation: Au+ —> Au^(3+) + 2e- and it worked out at the end but idk if I’m right

Charlyn Ghoubrial 2I

You would use the half reactions:
Reduction: Ni(OH)3(s) + e- —> Ni(OH)2(s)+ OH-(aq)
Oxidation: Cd(s) + 2OH-(aq) —> Cd(OH)2(s)+ 2e-

Charlyn Ghoubrial 2I

Your oxidation half reaction would start as Br- (aq) —> BrO3^(-) and then you would balance the oxygen by adding 3H2O(l) molecules on the left
Your reduction half reaction would start as O3(aq) —> O2(g) and then you would balance the oxygen by adding H2O(l) molecule on the right

Charlyn Ghoubrial 2I

I am a little confused too but read the textbook and try to do the examples they have within the sections and the hw problems and then go to workshops or step up sessions and office hours for clarifications. That’s what i do. Hope that helps

Charlyn Ghoubrial 2I

When the anode or cathode does not have a metal (s).
I don’t think we need to but I’m not entirely sure
And i think because one side is only aqueous so you’d need to add platinum as an electrode

Charlyn Ghoubrial 2I

If delta G is negative your reaction is spontaneous so the metal will dissolve
If delta G is positive your reaction is nonspontaneous so the metal will not dissolve

Charlyn Ghoubrial 2I

No you don’t include them

Charlyn Ghoubrial 2I

I think the half reaction would be Hg2^(2+) + 2e- —> 2Hg because that's the one that would give you +0.79 (i used appendix 2B to get that)

Charlyn Ghoubrial 2I

I understand how to look at the table to find out whether the reduction potentials are a larger positive number for the metal ion or water, but I am a little confused on where to go after that. If the cathode has a really large negative reduction potential(which we don't want), doesn't that mean th...

Charlyn Ghoubrial 2I

Ok so none of the replies mentioned the organic chemistry tutor! He’s really good and I find him better than khan academy but you won’t always find all the topics you need but he’ll provide the best and easiest explanations to the topics you search. Watch his videos if you need the basics to be expl...

Charlyn Ghoubrial 2I

It would decrease the concentration of Cr^3+ which decreases standard reduction potential of cathode so the cell potential will decrease ( Ecell= Ered-Eox)

Charlyn Ghoubrial 2I

The right side of the electrode is for reduction and the left side of the electrode is for oxidation so the half reactions are Oxidation (at anode): 2H2O —> O2 + 4H^(+) + 4e- Reduction (at cathode): O2 + 2H2O + 4e- —> 4OH- How do we know the oxidation reaction is H20 oxidizing to O2 and 4H+ when on...

Charlyn Ghoubrial 2I

Oxidation occurs at anode and reduction occurs at cathode so the half reactions are:
Reduction (at cathode): O2 + 4H+ + 4e- —> 2H2O
Oxidation (at anode):4OH- —> O2 + 2H2O + 4e-

Charlyn Ghoubrial 2I

The right side of the electrode is for reduction and the left side of the electrode is for oxidation so the half reactions are
Oxidation (at anode): 2H2O —> O2 + 4H^(+) + 4e-
Reduction (at cathode): O2 + 2H2O + 4e- —> 4OH-

Charlyn Ghoubrial 2I

Your 2 half reactions would be
Reduction (at cathode): 2(NO3)^(-) + 8H^(+) + 6e^(-) —> 2NO + 4H2O (Ecell= +0.96V)
Oxidation (at anode): 6Hg —> 3(Hg2)^(2+) + 6e^(-) (Ecell= -0.79V)

Charlyn Ghoubrial 2I

For b) oxidation occurs at anode and reduction occurs at cathode so then the reduction (at cathode) is O2(g) + 4H^(+) (aq) +4e^(-) —> 2H2O(l) and oxidation (at anode) is 4OH^(-) —> O2(g) + 2H2O(l) + 4e^(-)

Charlyn Ghoubrial 2I

605324529 wrote:Determine the standard cell potential:
2NO3(-)(aq) + 8H(+)(aq) + 6Hg(l) -->3Hg2(+2)(aq) + 2NO(g) + 4 H2O(l)

Thanks!

To find the standard cell potential, write the half reactions first then subtract the anode cell potential from the cathode cell potential and you should get +0.17 V

Charlyn Ghoubrial 2I

Don’t we only have to find the half reactions and balanced equations for this question? Do we need to calculate the cell potential as well?

Charlyn Ghoubrial 2I

I think you originally did it right because if you multiply the 2 by the reduction half reaction, you would get 2Au^(+) + 2e^(-) —> 2Au and then your 2e^(-) would cancel on each side giving you an overall balanced equation of 3Au^(+) —> Au^(3+) + 2Au

Charlyn Ghoubrial 2I

For the equations I got: MnO4^(-) + 8H^(+) + 5e^(-) —> Mn^(2+) + 4H2O and Fe^(2+) —> Fe^(3+) + e^(-)
So Mn would end up with a 2+ charge and Fe would end up with a 3+ charge if i did this correctly

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