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    im so confused on how you write redox equation from redox numbers . sulfur s reacts with concentrated nitric acid HNO3 to form sulfuric acid H2S04 ,nitrogen dioxide and water.construct an overall equation for this redox equation.
    the answer was S + 6HNO3 ---------------> H2SO4 + 6NO2 + 2H20
    i don't know how you get this equation
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    (Original post by sarikamannan)
    im so confused on how you write redox equation from redox numbers . sulfur s reacts with concentrated nitric acid HNO3 to form sulfuric acid H2S04 ,nitrogen dioxide and water.construct an overall equation for this redox equation.
    the answer was S + 6HNO3 ---------------> H2SO4 + 6NO2 + 2H20
    i don't know how you get this equation
    This is quite a skill to learn!

    First of all yo construct two half equations, one showing the species oxidised and the other the species reduced. You can only add H+ ions or water to balance the equation for atoms and electrons for charge.

    OK so far?

    So let's look at the sulfur. It changes from S to H2SO4

    Remember you can only add H+ ions or water:

    S + 4H2O --> H2SO4 + 6H+

    balanced for atoms but not for charge, so add 6 electrons to RHS

    Half-equation 1: S + 4H2O --> H2SO4 + 6H++ 6e

    OK so far?

    Now let's look at the nitric acid:

    Half-equation 2: HNO3 + H+ + 1e --> NO2 + H2O

    I balanced both atoms and charges at the same time ..

    NOW the essential step of making the electrons equal in both hanf equations. You can do this by mutiplying half-equation 2 by 6.

    Half-equation 2: 6HNO3 + 6H+ + 6e --> 6NO2 + 6H2O


    Now add the two half equations together and the electrons cancel out:

    Half-equation 1: S + 4H2O --> H2SO4 + 6H++ 6e
    Half-equation 2: 6HNO3 + 6H+ + 6e --> 6NO2 + 6H2O
    -------------------------------------------------------------------------------------------------- add

    S + 4H2O + 6HNO3 + 6H+ ---> H2SO4 + 6H+ + 6NO2 + 6H2O

    Now cancel out common terms from both sides:

    S + 6HNO3 ---> H2SO4 + 6NO2 + 2H2O

    ... and Robert is your Father's brother.
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    (Original post by charco)
    This is quite a skill to learn!

    First of all yo construct two half equations, one showing the species oxidised and the other the species reduced. You can only add H+ ions or water to balance the equation for atoms and electrons for charge.

    OK so far?

    So let's look at the sulfur. It changes from S to H2SO4

    Remember you can only add H+ ions or water:

    S + 4H2O --> H2SO4 + 6H+

    balanced for atoms but not for charge, so add 6 electrons to RHS

    Half-equation 1: S + 4H2O --> H2SO4 + 6H++ 6e

    OK so far?

    Now let's look at the nitric acid:

    Half-equation 2: HNO3 + H+ + 1e --> NO2 + H2O

    I balanced both atoms and charges at the same time ..

    NOW the essential step of making the electrons equal in both hanf equations. You can do this by mutiplying half-equation 2 by 6.

    Half-equation 2: 6HNO3 + 6H+ + 6e --> 6NO2 + 6H2O


    Now add the two half equations together and the electrons cancel out:

    Half-equation 1: S + 4H2O --> H2SO4 + 6H++ 6e
    Half-equation 2: 6HNO3 + 6H+ + 6e --> 6NO2 + 6H2O
    -------------------------------------------------------------------------------------------------- add

    S + 4H2O + 6HNO3 + 6H+ ---> H2SO4 + 6H+ + 6NO2 + 6H2O

    Now cancel out common terms from both sides:

    S + 6HNO3 ---> H2SO4 + 6NO2 + 2H2O

    ... and Robert is your Father's brother.

    THANKYOU ,
    you explained it so well and made it ALOT less confusing .
 
 
 
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