Oxydoreduction. HNO3 +H2S2+ - S(S) + NO(g)

Question

Oxydoreduction. HNO3 +H2S2+ -> S(S) + NO(g)

anonymous · Answer

$$HNO _{3} + H _{2}S ^{2+} -> S _{(s)} +NO _{(g)}$$

aaronq · Answer

$H_2S$ shouldn't have a charge

anonymous · Answer

oops, yeah but I still need to solve it haha

aaronq · Answer

so you have to balance the redox reaction?

aaronq · Answer

Follow this: Redox reactions: Half-cell method

1a. Assign oxidation states/numbers
2a. Separate (re-write) oxidation/reduction half-cell reactions

To individual half-cell reaction:
1b. Balance elements (ignore H and O; these are balanced with $H_2O$ later)
2b. Balance electrons transferred (i.e. change coeffients)
3b. Balance charges with: $\color{blue}{OH^-~in~base}$ or $\color{red}{H_3O^+~in~acid}$
4b. Balance H’s and O’s with $H_2O$
5b. Recombine half-reactions and cancel out redundancies.

anonymous · Answer

Hey! sorry for not responding... I never got notified. Anyway, thanks a lot! I never had proper list of how to do it so this really helps :)

Also just a question about "spectators" like how do you know when to NOT use part of the formula for example in 

SnCl_2 + KMnO_4 + HCl --> SnCl_4 + MnCl_2 + KCl + H_20

aaronq · Answer

No problem !

By spectators, you mean atoms not participating in the redox reaction?

If so, you just look for atoms that arent changing in oxidation state

$\sf \large \underbrace{SnCl_2}_{\normalsize (+2)(-1)}+ \underbrace{KMnO_4}_{ \normalsize  (+1)(+7)(-2)}+ \underbrace{HCl}_{\normalsize (+1)(-1) }→ \underbrace{SnCl_4}_{\normalsize  (+4)(-1)  }+ \underbrace{MnCl_2}_{\normalsize (+2)(-1)  }+ \underbrace{KCl}_{\normalsize    (+1)(-1)   }+ \underbrace{H_2O}_{\normalsize(+1)(-2)
 } $

Cl, K, H, O arent changing, so they are spectators.

anonymous · Answer

Okay! Thanks again! :)