ionic equation???

[albertlee]

I was given the task to write the balanced reaction of scandium fluoride reacting with calcium, but in ionic equation.

 

However, I dont know what is it, even when I looked up the answer.

 

it looks like

 

3Ca⁺ + Sc³⁺ ----> 3Ca²⁺ + Sc

 

what is it all about?? I mean, on LHS, why is Ca +?? isn't it 2+??

 

 

Albert

[albertlee]

Btw, please don't hesitate to ask me further details concerned with the complete question...

[mezarashi]

Ionic equations take a look at the ions and electrons interaction in a reaction, especially a reduction/oxidation reaction such as this one.

 

Indeed, could you please give me the original equation or question? I'm not sure why that calcium is a +1 myself. In its usual ionic form, it should be +2.

[Crash]

Calcium is this instance is +1, if it was +2 no reaction would take place. Im guessing this is just a hypothetical question posed by a high school chem teacher?

[albertlee]

Can you explain clearer, Crash??

 

what do you mean +2 the reaction wouldnot take place???

 

Albert

[BenSon]

What he means is that if the Ca was 2+ then it would not donate any electrons to the scandium and then you get no reaction.

 

~Scott

[DQW]

Yikes ! What a mess !

 

Crash/Benson : you are both wrong (as is the person that provided the OP with the quoted ionic equation). Where on earth do you see Calcium in a +1 oxidation state ? It only exhibits the +2 oxidation state in ionic form.

 

In any case, neither of these is what applies to the OP's question. I quote :

scandium fluoride reacting with calcium

The question is talking about Calcium metal - not Ca²⁺ and certainly not a fictitious Ca⁺.

 

In other words, the question reads :

[math]x~Ca + y~Sc^{3+} \longrightarrow~? [/math]

 

To the OP : Solve and balance the above reaction, using the fact that Calcium is more electropositive (more active) than Scandium. If you wish to convince yourself that this is true, look up an activity series table or a table of standard reduction potentials. Simply put, Calcium has a greater tendency to lose its valence electrons than Scandium, so it will simply dump its excess electrons onto Scandium, and make itself happy at the cost of the happiness of Scandium.

 

Incidentally, the mentioned reaction is the most commonly used method for the production of pure Scandium metal.

 

Also, albert, if you have a complete question, post the complete question. Holding back information does nothing but allow room for misinterpretation and confusion, and it could end up wasting the precious time put in by the responders, if an important piece of data were omitted. In general, not posting all the available details is extremely bad form. You are asking the question and others are doing you a favor by answering. The burden is on you to make the job easier for them, lest you tick them off for good.

 

Jdurg, Borek,.... where are you ?

[BenSon]

Yikes ! What a mess !

Crash/Benson : you are both wrong (as is the person that provided the OP with the quoted ionic equation). Where on earth do you see Calcium in a +1 oxidation state ? It only exhibits the +2 oxidation state in ionic form.

What a mess indeed, I'm seriously of my game.

 

~Scott

[albertlee]

"The main ore of scandium is thortveitite, Sc2Si2O7. This is converted into scandium fluoride, which reacts with calcium to produce scandium metal. Balance the ionic equation for the eaction between scandium fluoride and calcium:

...Ca^+ + ...Sc^3+ -----> ...Ca^2+ + ...Sc"

 

The above is the origininal question....

 

How would you do it??

 

Albert

[jdurg]

The original question is therefore incorrect and you should bring that up to whomever gave you the question. Ca+ simply does not exist in any stable compound used to produce scandium metal. Elemental calcium has a charge of 0 on it.

 

Even though I HATE when teachers/professors give hypothetical equations that are impossible to actually happen, I will help you out here.

 

The first thing to do is see what the oxidation numbers are for each of the species involved. On the reactants side you have Ca+ with an oxidation state of +1 (shudders) and Sc3+ with an oxidation state of +3.

 

On the products side, you have Ca2+ with an oxidation state of +2, and Sc with an oxidation state of 0. So now you have to split the products and reactants into half reactions. We'll start with just scandium.

 

If you remove the Calcium part from the equation, you get "Sc3+ -> Sc". In order to balance the charge, we have to add 3 electrons to the left side of the equation. This gives us "Sc3+ + 3e- -> Sc". The charge is now balanced. (Remember, when balancing equations both the charges and the atoms involved have to be balanced. To keep a charge balanced, just add and subtract electrons).

 

Now we'll look at the calcium. If we remove the Scandium portion, we get "Ca+ -> Ca2+". To balance the charge, we have to add an electron on the right side. This gives us "Ca+ -> Ca2+ + e-". Now we add the two equations together.

 

Sc3+ + 3e- -> Sc

Ca+ -> Ca2+ + e-.

 

We need to balance the number of electrons as we have three on the left side and only one on the right. So if we multiply the second equation by 3, we get the following:

 

Sc3+ + 3e- -> Sc

3Ca+ -> 3Ca2+ + 3e-

 

3Ca+ + Sc3+ -> 3Ca2+ + Sc

 

The electrons cancel each other out and you get the equation seen above.

[jdurg]

If the initial equation was a correct one, the answer would/should be:

 

3Ca + 2Sc3+ -> 3Ca2+ + 2Sc

[albertlee]

thanks Jdurg, but what do you mean that elemental calcium has a charge 0??

 

Isn't it 2+?? for pure calcium??

 

Secondly, Do you guys know what is the difference between Metal and Transition Metal??

The only thing I know is Transition Metal is more stable than Metal, but any more significant difference??

 

Apreciate any response

 

Albert

[albertlee]

If the initial equation was a correct one' date=' the answer would/should be:

 

3Ca + 2Sc3+ -> 3Ca2+ + 2Sc[/quote']

 

????

 

I think is wrong.

 

2ScF₃ + 2Ca ---> 2Sc + 2CaF₂

 

The above is the real reaction equation.

[DQW]

thanks Jdurg' date=' but what do you mean that elemental calcium has a charge 0??

 

Isn't it 2+?? for pure calcium?? [/quote']

Any stable element/compound must be charge neutral. So, for instance, CaO will have total charge 0, as will Ca (metal). In the first compound, the Ca-ion has a 2+ charge on it and the O ion has a 2- charge, making the whole thing have 0 charge. In the case of elemental Ca, it must have 0 charge to be stable by itself.

 

Undoubtedly, your textbook will have this covered. Give it a look-see.

 

Secondly, Do you guys know what is the difference between Metal and Transition Metal??

Transtion metals are those metals (all the elements, really) that form the d-block of the periodic table. Everything from Sc (top left) to Uub (yet to be named - atomic number 112) in the middle section of the periodic table, is a transition metal. They make up the pink blocks at http://www.webelements.com. Transition metals are characterized by having partially filled d-subshells (exceptions : Zn, Cd, Hg, Uub - which are sometimes considered transition metals and sometimes not). Other (non-transition) metals incluse the alkali and alkaline earth (Groups I,II; blue blocks - known as s-block elements), the Lanthanides and Actinides (green blocks - known as f-block elements) and the following p-block elements :Al, Ga, In, Sn, Tl, Pb, Bi. Their neighbors to the right are classified as semi-metals and those to their right are non-metals.

 

The only thing I know is Transition Metal is more stable than Metal, but any more significant difference??

Transition metals are themselves metals, so that sentence woudn't be right. However, if you said that transition metals are more stable than other metals, that would be often correct.

 

Another important characteristic of transition metals is that they form colored salts. The reason for the emission/reflection spectra of these salts being in the visible range (thats what it means when something is colored) is that these are the frequencies/energies associated with electronic transitions between two d-subshell energy levels (refered to as [imath]e_g~and ~t_{2g} [/imath] by spectroscopists) that form in a crystal. It is because of these transitions that these elements have this name.

[DQW]

????

 

I think is wrong.

No' date=' it is not. The last equation by jdurg is the correct equation for this reaction. Whether it is what your teacher wants is likely beyond the ability of anyone here to determine, but it is correct.

 

2ScF₃ + 2Ca ---> 2Sc + 2CaF₂

 

The above is the real reaction equation.

The above is NOT a balanced equation, and hence can not be real. Did you even count to see if there are equal numbers of F-atoms on both sides ? Someone is giving you bad information. I suggest you stop getting any more from the same source.

[albertlee]

thanks for pointing that out DQW, I just have bad days this few weeks that I consider a black period of mine this year, which is composed of a serie of unfortunate events. I can't believe why I have done the balanced equation wrongly since I was reasuring myself so much.

 

Nonetheless, I would ask my teacher again, and see if there's anything you guys might want to know.

 

Albert

[albertlee]

Finally, I have asked my chem teacher, and she said that she doesnot know how specifically that Calcium released its first electron in the 1st place, because she did not design the question herself, but however, she thinks that there is certain factor for exmple, pressure, that chemists apply on calcium to make it more "controllable" or less reactive, so calcium becomes only accepting to loose one more electon to the florine bit, to release its bond with scandium.

 

Maybe you guys have better explanation now??

 

Albert

[DQW]

Maybe you guys have better explanation now??

Not really ! The person who wrote down that first equation with Ca⁺ was just plain WRONG ! End of discussion.