ssNake's Chemistry [u2] Q's

[Lasercookie]

c=n/V

Not dried means that there's water left there. This could dilute the solution (therefore changing concentration).
Think of the role each equipment plays.

a. Pipette with water left in it - the sodium carbonate will be diluted, so less HCl needed to reach that point of change (can't remember the word for it). Remember that concentration is a ratio, so the HCl would have a higher concentration.

b. Flask with water left in it - we have an accurate calculation of sodium carbonate - a bit of extra water will not change the concentration. i.e. the total mole we pour in of the standard and the HCl won't be changed by the water. As such, the calculation for HCl won't be affected.

c. Burette - delivers the titre. Extra water here will dilute the HCl. So more HCl will be needed for that point of change. Therefore the concentration will be lower.

[REBORN]

Q - purpose of the electrolyte in a galvanic cell?
Q - how do we know when to use gas electrodes/what other special cases?

[Greatness]

The electrolyte dissolves the ions so that electricity can be conducted.
If the reactants arent solids then you would use gas/graphite etc electrodes. (Not too sure about this tho)

[REBORN]

Which of the following metals would be coated with lead when immersed in a solution of Lead (II) Nitrate?

a) Cu
b) Au
c) Sn
d) None

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I've forgotten which gets coated. But relating back to the original Daniel Cell - copper coated the copper electrode.

So the coating is done on the cathode?

[b^3]

The coating occurs at the cathode as it is undergoing reduction, it is gaining electrons. The metal ions need electrons to turn in the neutral atoms and become a solid.

To coat Pb, Pb²⁺ has to undergo reduction, i.e. the forward reaction. So Pb²⁺ only forms a negative gradient with Sn. So Sn will undergo oxidation, but be coated with Pb.

Answer: c) Sn.
EDIT: The answer could be d) none as well as the Sn corrodes, there would be nothing for the Pb to coat onto, it would form a solid, just not coat onto the rod that is being eaten away. I think I am overthinking this.

[REBORN]

The answer was C haha.

Q - A steady stream of oxygen was passed through an acidified solution of MgCl2. Write an eq for the reaction?

I have a feeling 'acidified' has some significance - and the O2? Hein [u2] does not help me.

[REBORN]

^ Anyone?

[Mao]

The answer was C haha.

Q - A steady stream of oxygen was passed through an acidified solution of MgCl2. Write an eq for the reaction?

I have a feeling 'acidified' has some significance - and the O2? Hein [u2] does not help me.

At the cathode, the reaction would be

At the anode, if the concentration of MgCl2 is high enough, the reaction would be

I think. (Not 100% sure. I don't have an electrochemical series handy)

[REBORN]

But what does 'acidified' solution mean?

How did you know to use 02+4H+ +4e --> 2H20 rather than O2 + 2H+ + 2e- ---> H2O2

Also Cathode = reduction right? But your oxidant is lower then the reductant (Cl is one above)....

I'm confused

[HenryP]

The "acidified solution" means that there are H+ ions present in the solution meaning that the reaction O2+4H+ +4e --> 2H2O can occur. Since the reaction O2+4H+ +4e --> 2H2O is higher up than O2 + 2H+ + 2e- ---> H2O2 it is the one that is more likely to occur.

Mao specified "if the concentration of MgCl2 is high enough" meaning that the concentration may be higher than 1M. If this is the case, the order in the electrochemical series may not apply as it is not at standard conditions and the reaction may still be spontaneous even though as you said the Reductant is higher than the Oxidant.
Hopefully this explanation is correct and cleared things up for you.

[REBORN]

Oh so under 1M normal conditions there is no reaction?

[HenryP]

Well it may still happen as there are literally next to each other on the electrochemical series, but if you strictly stick to the electrochemical series then yes, no reaction will occur.

[REBORN]

I understand

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Lawn fertiliser contains ammonium ions (NH4+). A 1.234 g sample of lawn fertiliser was dissolved in water to make a 250.0 mL solution. A 20.00 mL aliquot of this solution was added to a flask containing 20.00 mL of 0.1022 M sodium hydroxide
solution. The flask was heated until the reaction:NH4 +(aq) + OH–(aq) → NH3(aq) + H2O(l) was complete. Excess sodium hydroxide in the resulting solution was titrated with 0.1132 M hydrochloric acid, using phenolphthalein as indicator. The end point was reached when 9.97 mL had been added. Calculate:
a the amount, in mol, of HCl used in the titration
b the amount of NaOH in excess after reaction with the fertiliser
c the amount of NaOH that reacted with the NH4+ ions
d the amount of NH4+ ions in the 1.234 g fertiliser sample
e the percentage by mass of nitrogen in the fertiliser, assuming nitrogen is only present as ammonium ions.

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I can do calculations/stoich etc etc....I just need someone to explain to me what's going on and how to use the 2 equations with the limiting and excess and all that shiz.

[illuminati]

HCl + NaOH -----> NaCl + H2O
This reacts in a 1:1 ratio
Therefore: n(HCl) = n(NaOH) excess
n(NaOH) reacted with NH4+ = n(NaOH)initial – n(NaOH)excess
NH4+ + OH- ----> NH3 + H2O
Again, 1:1 ratio
Therefore n(NaOH) reacted with NH4+  = n(NH4+)

(It's a back titration, Unit 3 chem)

[REBORN]

Classify this reaction: HCN + H20 -----> CN- + H3O+

I said 'acid-base'

Checkpts says 'ionisation'

They're both right?