Question

Trying to solve a problem regarding the Half-Reaction method, more information posted below.

Answer 1

"What are the coefficients of the reactants and products in the balanced equation above?"

\[\sf SO_{4}^{2-}(aq)+Sn^{2+}(aq)+(???) \rightarrow H_{2}SO_{3}(aq)+Sn^{4+}(aq)+(???)\]

Answer 2

The blanks should be H2O and positive Hydrogen ions.

Answer 3

@Hoslos , let me know if you get a chance; I would definitely appreciate help.

Answer 4

I'm unsure of both how I'm supposed to determine where H2O and H+ go, and how to determine the stoichiometric coefficients in this case; what exactly is the procedure to solve this kind of a problem?

Answer 5

Is it H+ in the first side and H20 in the RHS?

Answer 6

That's the thing, I'm not sure. How do I determine which goes where? (The H20 and H+ are being added to the reaction to help balance it somehow)

Answer 7

Sn2+(aq)→Sn4 + 2 e- (oxidation)
2e-+ 4H+ + SO4−2(aq)→H2SO3(aq)+H2O (reduction)
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4H+ + SO4−2(aq)+Sn2+(aq)→H2SO3(aq)+Sn4+(aq)+H2O

Answer 8

I got a ton of question marks and undisplayed symbols.

Answer 9

reload the page

Answer 10

You got it. I WAs just seeing in my redox electrod potentials table.

Answer 11

One sec, reloading

Answer 12

Now, all you have to do to make a unique equation, you need to make sure that one equation has electrons in the LHS and the other on the RhS. If this is not the case, than you will have to swap one of them. Before you combine them, the best way to equate is to first check if the number of electrons is the same for both equation. If not, you will have to equate them, making sure that they both get the same number of electrons.

Answer 13

(Sorry, trying to eat at the same time, heh)-alright, I'm still just reading over what Cuanchi posted, trying to make sure I 100% understand before moving on

Answer 14

According to your final equation, it is balanced. Maybe what confuses you is the oxidation number. THAT, you never balance. It is just the number of atoms within the element.

Answer 15

According to your second ionic equation, i have a feeling that it might be wrong. RE-checket.

Answer 16

I only put up one equation, lol Cuanchi did everything else, and I'm not sure I follow what he did. I only put up one.

Answer 17

I meant Cuanchi's 2nd equation.

Answer 18

@Cuanchi , just letting you know, using the LaTeX editor makes things way clearer, heh

Answer 19

Anyway, Bias, how did you get your equation ?

Answer 20

Which one, the one I originally posted?

Answer 21

The one in the prompt was given, that being the only one I posted. It was given.

Answer 22

Yes.

Answer 23

http://i.imgur.com/FhYcczr.png

Answer 24

Yes, so I believe Cuanchi 's final equation is correct. As to balance it, I explained earlier. DID YOU Read it and understood it ?

Answer 25

I think I'm figuring it out.

Answer 26

Yes, Cuanchi 's final equation is correct. NOW re-read how to make sure the equation is right.

Answer 27

I don't really understand the way you're *trying* to tell me how to solve this problem, but I think I've figured it out from looking to another source and am going to try to post my workings here. Splitting the original equation into its two unbalanced Half-Reactions:\[\sf SO_{4}^{2-}\rightarrow H_{2}SO_{3}\]\[\sf Sn^{2+}\rightarrow Sn^{4+}\]

Answer 28

\[Sn ^{2+} \rightarrow Sn ^{4+}+ 2e ^{-}\]
\[2e ^{-}+4H ^{+}+SO _{4}^{2-}\rightarrow H _{2}SO _{3}+H2O\]
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\[4H ^{+}+SO _{4}^{2-}+Sn ^{2+}\rightarrow Sn ^{4+}+H _{2}SO _{3}+H2O (l)\]

Answer 29

\[\sf SO_{4}^{2-}+2H^{+}\rightarrow H_{2}SO_{3}\]Not sure how to account for the additiona O on the left, using the Half-Reaction method.

Answer 30

Alright, so we're allowed to add water molecules on the right, makes sense.

Answer 31

for the Sn equation you balance the charges with 2 electrons at the right ( it is an oxidation, the mass of the elements are balanced

Answer 32

the second half equation you balance the O with H2O, then the H with H+ in the other side of the equation and finally the charge with 2 e in the left side of the equation (this is the reduction always the electrons in the reduction are in the reactants side)

Answer 33

Bias, actually, that equation is SO-4 +4H+ + 2e - SO2 + H2O, that is why the SO2 + H2O can be the same as H 2SO3.

Answer 34

Alright, just going to look over this for a moment, thanks, guys.