Recent content by 4LeafClover

I understand

The Ag+ will react with Cl- and form AgCl. So less Cl- means more Ag+

So how does a higher concentration of AgCl make for a lower concentration of Cl-? Please just explain

Well I guess the difference between the two would be that in the second, Ag is reduced while the oxidation state of Cl- remains the same. The concentration for this equation is given in terms of Cl- instead of the species that is reduced whereas in the first instance the concentration is given...

For Ni: Anode: Ni(s) --> Ni^2+(aq) + 2e- Cathode: Ni^2+(aq) + 2e- --> Ni(s) For Ag: Anode: Ag(s) + Cl-(aq) --> AgCl(aq) + 1e- Cathode: AgCl(aq) + 1e- --> Cl-(aq) + Ag(s)

Ok, well, if the reaction is AgCl(s) + e- = Ag(s) + Cl-(aq) and the two cell compartments have [Cl-]= 1.41×10−2 M and [Cl-]= 2.70 M, that means that there is higher electron concentration in the first because lower Cl- concentration means that there are more free electrons. Is that what they...

Actually my textbook is telling me that the more concentrated solution is the cathode but my homework is telling me the more dilute ([Cl-]= 1.41×10−2 M) is my cathode. Can you explain to me what you mean because just telling me I'm not understanding isn't helping me to understand.

My chemistry book explicitly states that, in a concentration cell whose compartments both consist of a strip of nickel metal immersed in solutions of different concentrations of Ni^2+, that oxidation occurs in the half cell containing the more dilute solution and is therefore the anode and that...

Ya, I thought my calculations were correct too, but they're not for some reason. And what I meant by more concentrated, I wondering if, when you make a concentration cell and you start off with one solution that is more concentrated and then one that is more diluted (in order to make electrons...

Question: A voltaic cell is constructed with two silver-silver chloride electrodes, each of which is based on the following half-reaction: AgCl(s) + e- = Ag(s) + Cl-(aq). The two cell compartments have [Cl-]= 1.41×10−2 M and [Cl-]= 2.70 M, respectively. What is the emf of the cell for the...

Wow, thanks! Hahaha, that was the most frustrating problem I've dealt with in a while, but it feels very good to have gotten through it. Thanks for all the help!

Wow, yes, I think I finally get it. I am going to have to go back and look at all we've talked about a couple of times though to make sure I've got it down for exams and such. So since .10 moles of H2PO4- is my acid and the 0.35 moles of HPO4^2- is my base, my equation is going to be the...

H2PO4- + NaOH --> H2O + HPO4^2- So they react in a 1:1 ratio. We produced 0.45 moles of H2PO4- in the reaction between NaOH and the first proton of H3PO4. Therefore, we have 0.45 moles of H2PO4- to work with and 0.8 - 0.45 = 0.35 moles of NaOH left to work with. Since they react in a 1:1...

0.175 moles of H2PO4- will be neutralized. There will be 0.275 moles H2PO4- left. This will produce 0.175 moles of HPO4^2-. Extra Credit: If there were still unreacted NaOH left, it would neutralize as much HPO4^2- as possible - until excess base ran out.

0.45 moles of NaOH will react with the first proton of H3PO4. After the first proton is neutralized, there will be H2PO4- present. The excess base will neutralize as much H2PO4- as possible until all NaOH is used up.