ssNake's Chemistry [u2] Q's

[REBORN]

What is a spontaneous redox reaction and how do we predict it using the electrochemical series?

The q: Silver metal is placed in a copper nitrate solution. Is it a spontaneous redox reaction?

How do I use my series? :S

[tony3272]

A spontaneous redox reaction is one which occurs naturally. i.e the electrons want to move from one thing to the other.
Since the silver metal is going to be oxidised, the reaction will occur if it is a stronger reductant. However, if you look on you electrochemical series, you will see that silver metal is higher up than copper. It is a stronger oxidant and hence the reactio will not be spontaneous.

The simplest way to use your series is to draw a line between the two reactants. If the gradient of the line is negative, the reaction is spontaneous. If it's positive, then it is not spontaneous.

[scocliffe09]

A spontaneous redox reaction is one which occurs naturally. i.e the electrons want to move from one thing to the other.
Since the silver metal is going to be oxidised, the reaction will occur if it is a stronger reductant. However, if you look on you electrochemical series, you will see that silver metal is higher up than copper. It is a stronger oxidant and hence the reactio will not be spontaneous.

The simplest way to use your series is to draw a line between the two reactants. If the gradient of the line is negative, the reaction is spontaneous. If it's positive, then it is not spontaneous.

i.e. oxidant (LHS) must be higher on the series than reductant (RHS). A common method for doing this is to draw a 'backwards Z', which will show you which half-equations are occurring as well.

[REBORN]

FeCl3 + 3NaOH ---> Fe(OH)₃ + 3 NaCl

You have 0.02mol FeCl3
You have 0.018 mol NaOH

Please work out the mass of iron(III) hydroxide precipitated!

[Russ]

NaOH is the limiting reagent and FeCL3 is in excess
therefore we use .018 in our calculations
stoich ratio is 1/3 (what we want to find out / what we already know)
n(Fe(OH)3) = .018/3 = .006
m(Fe(OH)3) = .006 * 106.9 = .6461g

oh wow, i haven't done chem in ages :S

[REBORN]

A student is given a mixture of calcium chloride and calcium nitrate weighing 2.365g and told to determine the mass of each in the mixture. He adds 50mLs distilled water to the mixture and when it is totally dissolved he added an excess of silver nitrate to precipitate out all the chloride ions. At the end of the reaction it is found that the mass of silver chloride was 3.524g. Use this info to determine the mass of calcium chloride and calcium nitrate int he original mixture.

---

Lol? Am I meant to know this :/

Workings out would be appreciated!

[tony3272]

I'm not sure about year 11, but for year 12 we do this in gravimetric analysis. Basically the reaction between Calcium chloride and silver nitrate is:


using n=m/Mr  for silver chloride, you get n=0.024575mol

using your mole ratio, 0.012287 mol of Calcium chloride reacted. 

Using this information we can calculate that the mass of calcium chloride was 1.365g
and therefore the mass of calcium nitrate was 2.365g - 1.365g =1.000g



EDIT: Please correct me if i'm wrong. i'm having a bad day today 

[REBORN]

It looks so blindingly obvious when you type it out for me

thanks.

[Russ]

Worded questions exist to try and confuse people. You can usually bypass this by circling the chemicals used and their amounts and ignoring everything else

[REBORN]

Redox eq given:

2Fe(s) + O₂ (aq) + 2H₂O (l) ----> 2Fe2+ (aq) + 4OH- (aq)

Q: Write the two half-eqs.

My thinking

Assign oxidation numbers.

2Fe(s) +             O₂ + 2H₂O (l) ----> 2Fe2+ (aq) + 4OH- (aq)

0      0? free element?     +1 -2          ----->    +2              lost here..charges don't add

[tony3272]

The oxidation numbers do add.
On the left hand side, nothing has a charge and so the total oxidation sum is 0, and on the left hand side oxidation sum is 2 times (2+) + 4 times (1-), which cancels to zero.

Your half equations will be

Fe(s) ----> Fe²⁺(aq) + 2e^-

O₂(aq) + H₂O +4e^- -----> 4OH^-

[REBORN]

Fe makes sense as it went from 0 to +2 which is oxdiation/increase in oxidation number.

I don't get the reduction eq from the overall eq.

We have O₂ (aq) with NO CHARGE? so 0 oxidation number and H2O with O having an oxidation number of -2.

Then we look at the right hand side and 4OH-....the oxidaton number of O is -2.

So O2 has been reduced from 0 to -2? Or do we have to take into account the oxidation number of H2O?

Also there should be a 2 infront of the H20 ...otherwise your H's aren't balanced.

[tony3272]

We just look at the O₂, as it's those two oxygens which go from 0 to -2.  If we look at all 4, then we would need to add 8 electrons rather than 4.

[REBORN]

A commercial concrete cleaner contains concentrated HCl. A 25mL volume of cleaner was diluted to 250mL in a volumetric flask. A 20mL aliquot of 0.448M Na₂CO₃ was placied in a conical flask. Methyl orange indicator was added and the solution was titrated with the diluted cleaner. The methyl orange indicator changed permanently from yellow to pink when 19.84 mL of the cleaner was added. Calculate the conc of HCl in the concrete cleaner.

So...

The eq is 2HCl + Na₂CO₃ ----> 2NaCl + CO2 + H20

n(Na₂CO₃) is 0.448 * 0.02 = 0.00896mol.

Thus HCl is twice that according to eq which is 0.01792mol.

And now?

[b^3]

So the volume this HCl is in is the volume of the cleaner you added (as it is what is in the cleaner). So divide the mol of HCL by the volume (in L) so  0.01792mol/0.01948L to give the concentration in M. This will be the concentration of the dilute cleaner. So times it back by the dilution factor, 250/20 which will give the conc of the cleaner.

N(HCl)=0.01792mol
so c=n/v=0.01792/0.1948
=0.09199M
that is in the diluted sample
so in the inital sample the concentration will be 250/20 *0.09199
=1.150M

EDIT: Fixed it