andrew's really crappy chem thread

[MagicGecko]

theres no 'HCl' in the answer... (for some odd reason)

The reason why HCl isn't in the answer is because the H+ cancel out. When you are writing ionic equations you split them into their  relative ions. Take Yacoubbs balanced equation for HCl:
2HCl --> Cl₂ + 2H⁺ + 2e
Now you split the HCl into its ions and get:
2H⁺ + 2Cl^- --> Cl₂ +2H⁺ + 2e
The H+ cancel each other out therefore giving you:
2Cl^-(aq) --> Cl₂(g) + 2e

Hope I cleared that up

[soNasty]

Knew it! Thanks yacoubb and gecko

[soNasty]

Can someone explain how I'd write a precipitation reaction for

Pb(NO3)2 + KI

and

(NH4)2S + Cd(NO3)2

[Yacoubb]

Can someone explain how I'd write a precipitation reaction for

Pb(NO3)2 + KI

and

(NH4)2S + Cd(NO3)2

Pb(NO3)2 (aq) + 2KI(aq) --> PbI2(s) + 2KNO3
Iodine is a halide; any silver, lead or mercury cation that reacts with a halide anion forms a precipitate. Thus, the lead iodide is the precipitate that forms.

(NH4)2S(aq) + Cd(NO3)2(aq) --> 2NH4NO3(aq) + CdS (s)
Any cation sulfide (except if the cation is an ion of an element in group 1- remember this) will form a precipitate. This also applies to ___ carbonate, hydroxide, oxide and phosphate.

[soNasty]

How is the oxidation number of V in  +5?

[Yacoubb]

How is the oxidation number of V in  +5?

X-4=2
X= +6

V should have an oxidation number of +6.

[soNasty]

Where did you get -4 from?
This is from A+ chem unit 3 q12

[Yacoubb]

Where did you get -4 from?
This is from A+ chem unit 3 q12

Oxygen has a -2 charge. Because the VO2 molecule contains 2 oxygens, the charge becomes -4.

[soNasty]

It doesn't, it's just VO with a charge of 2+

[Yacoubb]

It doesn't, it's just VO with a charge of 2+

OMG I completely misread it!

Okay

x-2=2+
X=4+

[soNasty]

Aha yeah that's what I thought... For some reason it's saying that it's +5.. My periodic table says Vanadium is +5 too.. Oh well

[lzxnl]

Vanadium is a transition metal, element number 23 (if I remember correctly) with the electron configuration [Ar] 3d³4s². As it is a transition metal, it can take multiple oxidation states from +0 (in metallic form) to +5, where all of its 4s and 3d electrons are lost as in V2O5. The periodic table only shows how many valence electrons vanadium has, not its only oxidation state. Be careful. Transition metals can retain valence electrons when forming ions. VO²⁺, for instance, highlights this. When working out these oxidation numbers, assume nothing about the metal if possible.

[soNasty]

The iron content in a 0.200 g sample of fencing wire was determined by dissolving the wire in dilute sulfuric acid and making up the resulting pale green solution of Fe2+ ions to 25 mL. The solution was titrated with 0.0300 M potassium permanganate (KMnO4) solution, which is purple in colour. A titre of 20.22 mL was obtained. The solution of Mn2+ and Fe3+ ions produced by the reaction was almost colourless.

How would i go abouts in writing a titration reaction for this?

[nhmn0301]

The iron content in a 0.200 g sample of fencing wire was determined by dissolving the wire in dilute sulfuric acid and making up the resulting pale green solution of Fe2+ ions to 25 mL. The solution was titrated with 0.0300 M potassium permanganate (KMnO4) solution, which is purple in colour. A titre of 20.22 mL was obtained. The solution of Mn2+ and Fe3+ ions produced by the reaction was almost colourless.

How would i go abouts in writing a titration reaction for this?

Fe 2+ -> Fe 3+. This is an oxidation reaction.
Fe2+ -> Fe 3+ + e ( you need to include states, but I'm too lazy to type here )
MnO4(1-) -> Mn 2+ ( K just acts as a spectator ions, that's why I don't write in down, but looking at oxidation number, you know this is reduction)
MnO4 (1-) + 8H+   5e     -> Mn 2+ +  4H2O
You need to add these 2 half equations together. Hence, you need to make their electrons cancel each other out.simply multiply the equation of Fe 2+ by 6.
Overall equation:
5Fe (2+). + MnO4 + 8H+ -> 5Fe (3+) + Mn (2+) + 4H2O ( please include states)
Hope this helps!

[soNasty]

hey thanks! how would i deduce the fact that K is a spectator ion in this case?