[JadenErius]

Hey everyone, can someone give me the half equations for this reaction

6H2S + 2HNO3 -> 6S + 2NO + 4H20

[cliverlong]

Hey everyone, can someone give me the half equations for this reaction

6H2S + 2HNO3 -> 6S + 2NO + 4H20

What are the oxidation states of S and N on left and right-hand side of equation?

What does this tell you of the electron gain/loss of each species?

From this write the half-equations

Hint: oxygen in compounds almost always has oxidation state of 2. In all neutral molecules the oxidation states add up to zero. If you haven't studied oxidation states I doubt you can tackle the question.

Clive

[JadenErius]

sulphur = 0 Nitrogen = 2
I am studying oxidation states but this particular question stuns me and my chemistry professor. Is nitric acid a common reducing agent?

[Schrödinger]

sulphur = 0 Nitrogen = 2

Those are the oxidation states of S and N on the RHS(right hand side). What about those on the LHS? Is there any change in oxidation state that you observe? Which species is getting reduces and which is getting oxidized?

[JadenErius]

not mistaken the oxidation state of nitrogen on the LHs is +5 and sulphur is -2. Hydrogen sulphide is oxidised to sulphur and nitric acid is reduced to nitrogen monoxide. But what is the half equation for the reduction of nitric acid

[cliverlong]

not mistaken the oxidation state of nitrogen on the LHs is +5 and sulphur is -2. Hydrogen sulphide is oxidised to sulphur and nitric acid is reduced to nitrogen monoxide. But what is the half equation for the reduction of nitric acid

Ok here's a starter

the "central" change for nitrogen is from

(2) NO₃^- to (2) NO

So oxygens are being lost. Where have they gone? Well the hydrogen ions "soak up" the oxygens to form water

BTW, your original equation isnt balanced. You have 14 H on left and 8 H on right. thsi will make getting correct ionic equations tricky (impossible)

Clive

[jk22]

Hallo, nice to meet you. I'm new to this forum.

Hey everyone, can someone give me the half equations for this reaction
6H2S + 2HNO3 -> 6S + 2NO + 4H20

can we write, that :

LSH : 6H₂S+2HN⁽⁺⁵⁾O₃^-->12H⁺+6S^-2+2H⁺+2NO₃^-
->14H⁺+6S^-2+2N⁽⁺²⁾O-6e^-+4O^-2
->4H₂O+2NO+6S^-2+(6H⁺-6e^-)
->4H₂O+2NO+6S+6(H⁺+e^-)
->6S+2NO+4H₂O+(3H₂) X ?

but maybe it could be written (to be experimentally checked), that 1electron, or negative charge, would become 1 anti-proton : 1e^-->1p^-, which would annihilate the H+ ion :

->4H₂O+2NO+6S+6(H⁺+p^-) (lack of mass balancing, or energy : - E=6(m_p-m_e)c²) ?

->6S+2NO+4H₂O (-E,  and hence an endothermic reaction) ?

cya.

PS.: could this be used to build a "chemical temperature regulator" ? (It were welcomed by such a hot summer)

[JadenErius]

Hallo, nice to meet you. I'm new to this forum.

Hey everyone, can someone give me the half equations for this reaction
6H2S + 2HNO3 -> 6S + 2NO + 4H20

can we write, that :

LSH : 6H₂S+2HN⁽⁺⁵⁾O₃^-->12H⁺+6S^-2+2H⁺+2NO₃^-
->14H⁺+6S^-2+2N⁽⁺²⁾O-6e^-+4O^-2
->4H₂O+2NO+6S^-2+(6H⁺-6e^-)
->4H₂O+2NO+6S+6(H⁺+e^-)
->6S+2NO+4H₂O+(3H₂) X ?

but maybe it could be written (to be experimentally checked), that 1electron, or negative charge, would become 1 anti-proton : 1e^-->1p^-, which would annihilate the H+ ion :

->4H₂O+2NO+6S+6(H⁺+p^-) (lack of mass balancing, or energy : - E=6(m_p-m_e)c²) ?

->6S+2NO+4H₂O (-E,  and hence an endothermic reaction) ?

what the heck? anti protons? Isnt that a bit out of context here?


anyway, if i am not mistaken actual equation should be
3H2S + 2HNO3 -> 3S + 2NO + 4H2O

Is the ionic equation for reduction of nitric acid this

6H+ + 2HNO3 -> 2NO + 8H20

If so where does the hydrogen ions come from (if from the hydrogen sulphide what causes the hydrogen sulphide to be reduced if the nitric acid requires hydrogen ions to be reduced. If i am making any sense)

[jk22]

Hello, nice to meet you.

yes, I was trying to keep the initial equation, and use physics instead (out of context, and moreover it lacks of energy), but in this case it's more reasonable to modify the initial data.

thx.

[JadenErius]

err i dont think that helps solve this particular problem