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Chemistry 15 Online
OpenStudy (anonymous):

Hi! Can anyone help me out with a small redox reaction?

OpenStudy (anonymous):

\[S2H2+HN2O2 \rightarrow (S2O6)-2 +NH2OH\]

OpenStudy (anonymous):

How can I calculate the oxidation number for N? H is +1, and O is -2 time 2 so it's -4, is it possible the Oxidation number for N will be 1.5?!

OpenStudy (anonymous):

@iGreen

OpenStudy (anonymous):

Maybe there's a mistake in the question?

OpenStudy (igreen):

Errr...I have no idea..

OpenStudy (anonymous):

Thanks though!

OpenStudy (anonymous):

@JFraser What do you think?

OpenStudy (jfraser):

ill have to look at it for a minute

OpenStudy (anonymous):

thank you!

OpenStudy (jfraser):

if that formula is correct, the oxidation of the nitrogen in the reactant would have to be 1.5, which doesn't happen. I'd suggest that the formula in the reactant be \(HNO_2\), which is nitrous acid.

OpenStudy (anonymous):

I knew it! Thanks allot! and one more clarification if I may ask, \[A: Ca^+2 +2e \rightarrow Ca (E= -2.87) or B: Rb^+2 +2e \rightarrow Rb (E= -2.925)\]

OpenStudy (anonymous):

which one is a deuction cell and which one is the oxidizing cell?

OpenStudy (anonymous):

reduction

OpenStudy (jfraser):

i can't answer that with the information you've provided. Is there more?

OpenStudy (anonymous):

no, that's all

OpenStudy (anonymous):

they say that both of them are reduction reactions

OpenStudy (anonymous):

but which of them is a reduction half cell and which one of them is an oxidizing half cell?

OpenStudy (anonymous):

it doesn't make sense to answer on it without more details, is it?

OpenStudy (anonymous):

*does it

OpenStudy (anonymous):

@JFraser it's just gen chem but they endeavor to make hard life for us lol

OpenStudy (jfraser):

both reactions are listed as reductions, that's true

OpenStudy (jfraser):

but when they are \(combined\), one of them will be forced to "run" backwards and be an oxidation

OpenStudy (anonymous):

right! but how can you tell which one of them? the answer is that A will be a reduction half cell and B will be an oxidation half cell

OpenStudy (anonymous):

but how am I supposed to figure that out? by the E?

OpenStudy (jfraser):

you look at the E values they give, which are the \(reduction\) potentials. The LARGER reduction potential will stay a reduction. The SMALLER reduction potential will be the reaction that flips backwards and will be the oxidation

OpenStudy (anonymous):

so it's a negative E and hence A will go through reduction since it's bigger compared to B?

OpenStudy (anonymous):

is that the reason?

OpenStudy (anonymous):

@JFraser am I right?

OpenStudy (jfraser):

A has a larger reduction potential than B, so when you pair them together, A will be the reduction and B will be flipped and be the oxidation. Yes

OpenStudy (anonymous):

Thank you!!!! You are brilliant!!!!!

OpenStudy (jfraser):

YW

OpenStudy (anonymous):

I hope one day to help in return

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