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    Sulphide ions (S^{2-}) react with conc nitric acid to form NO and S.

    They made me draw a chart of the electrode potentials of S (-0.48v), NO3(+0.8v) and N2O4(+1.03v). So obviously S reduces NO3, and NO3 reduces N2O4 right?, or does S reduce N2O4 too?

    Then it asks, the reaction takes place in 2 steps. Use electrode potential chart to calculate Ecell for each of the 2 steps. Write the overall balanced equation for the reaction.

    I can merge 2 half equations, for each one, but i can't do the part in bold.

    +REP for good explanation!

    Thanks
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    Could you post the full half-equations please? Makes it easier to visualise what is happening in the cell (for me at least :P, perhaps someone else can answer without the half equations).

    But to merge half equations, you just make the number of electrons in each equal. For example:

    X + 2e ----> X2-
    Y-1 ----> Y + e

    For the above equation you would multiply the Y half equation by two, to get:

    X + 2e ----> X2-
    2Y-1 ----> 2Y + 2e
    And hence the overall equation will be:
    2Y-1 + X ----> X2- + 2Y
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    what board are u mos def ?

    if ur edexcel...i might die :p:
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    (Original post by Wenzel)
    Could you post the full half-equations please? Makes it easier to visualise what is happening in the cell (for me at least :P, perhaps someone else can answer without the half equations).

    But to merge half equations, you just make the number of electrons in each equal. For example:

    X + 2e ----> X2-
    Y-1 ----> Y + e

    For the above equation you would multiply the Y half equation by two, to get:

    X + 2e ----> X2-
    2Y-1 ----> 2Y + 2e
    And hence the overall equation will be:
    2Y-1 + X ----> X2- + 2Y
    lol sorry, anyway i know how to merge 2, i need to know this though:

    S + e^- ==> S^{2-} = -0.48v

    2NO_3^- + 4H^+ + 2e^- ==> N_2O_4 + 2H_2O = +0.80v

    N_2O_4 + 4H^+ + 4e^- ===> 2NO +2H_2O = +1.03v

    Merging any 2 is fine, i can do that. But how would i give the overall reaction corresponding to the original question?

    The answer is: 3S^{2-} +2NO3^- + 8H^+ ==> 2NO + 3S + 4H_2O

    +REP!!!! COME ONNN lol
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    Well N2O4 is a powerful oxidant; 2N2O4 + Zn ---> 2 NO + Zn(NO3)2
    so then maybe the first step is ZnS + 2HNO3 ---> N2O4 + S + Zn + H2O??

    anyway the point is that the intermediate N2O4 will not appear in the overall reaction as it is formed and then destroyed
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    oh now you post the half equations ...damn
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    (Original post by EierVonSatan)
    oh now you post the half equations ...damn
    lol sorry it took ages to write em up on latex, can you show me now pleeeease?
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    no they were rough guesses (i added Zn as counters, and i didnt account for 'magic' protons!) its now 3S2- + 2NO3- + 8H+ ---> N2O4 + 2H2O + 4H+ + 3S -----> 2NO + 4H2O + 3S

    so yeah the intermediate N2O4 is made and consumed so you can just cancel them out
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    (Original post by EierVonSatan)
    no they were rough guesses (i added Zn as counters, and i didnt account for 'magic' protons!) its now 3S2- + 2NO3- + 8H+ ---> N2O4 + 2H2O + 4H+ + 3S -----> 2NO + 4H2O + 3S

    so yeah the intermediate N2O4 is made and consumed so you can just cancel them out
    Wait, but how did you do that?
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    (Original post by Mos Def)
    lol sorry, anyway i know how to merge 2, i need to know this though:

    S + e^- ==> S^{2-} = -0.48v

    2NO_3^- + 4H^+ + 2e^- ==> N_2O_4 + 2H_2O = +0.80v

    N_2O_4 + 4H^+ + 4e^- ===> 2NO +2H_2O = +1.03v

    Merging any 2 is fine, i can do that. But how would i give the overall reaction corresponding to the original question?

    The answer is: 3S^{2-} +2NO3^- + 8H^+ ==> 2NO + 3S + 4H_2O

    +REP!!!! COME ONNN lol
    I'm guessing the S half equation should be 2e. If so then the final equation is worked out in the way I said earlier, more or less. First combine the second two equations:

    2NO3- + 4H+ + 2e- ==> N2O4 + 2H2O = +0.80v
    N2O4 + 4H+ + 4e- ===> 2NO +2H2O = +1.03v
    Like EierVonSatan said, the N2O4 will cancel. So you get:

    2NO3- + 8H+ + 6e ----> 2NO + 4H2O

    Then just balance the sulphur equation and combine the two.

    3*(S + 2e ---> S2-)
    2NO3- + 8H+ + 6e ----> 2NO + 4H2O
    3S2- + 2NO3- + 8H+ ---> 2NO + 4H2O + 3S
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    its akin to A + 5B = 4C + D + A ... A is on both sides so you can cancel so overall 5B = 4C + D
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    (Original post by Wenzel)
    I'm guessing the S half equation should be 2e. If so then the final equation is worked out in the way I said earlier, more or less. First combine the second two equations:

    2NO3- + 4H+ + 2e- ==> N2O4 + 2H2O = +0.80v
    N2O4 + 4H+ + 4e- ===> 2NO +2H2O = +1.03v
    Like EierVonSatan said, the N2O4 will cancel. So you get:

    2NO3- + 8H+ + 6e ----> 2NO + 4H2O

    Then just balance the sulphur equation and combine the two.

    3x(S + 2e ---> S2-)
    2NO3- + 8H+ + 6e ----> 2NO + 4H2O
    3S2- + 2NO3- + 8H+ ---> 2NO + 4H2O + 3S
    Did you flip around the Sulphur half equation at the end, and why if you did?
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    (Original post by Mos Def)
    Did you flip around the Sulphur half equation at the end, and why if you did?
    In the overall equation you mean? Yeah.

    Basic reason why: the question says you're adding the S2- ions to the conc nitric acid (NO3-), so it had to be swapped around (since S2- and NO3- are the reactants). Remember all redox potentials are in equillibrium, and are reversible.

    Other reason: since the sulphur potential is the more negative of the two, equilibrium will lie to the left. The equilibrium for the nitric acid will lie to the right, since it is the more positive of the two potentials. This means the NO3- will be reduced by S2- (and the S2- oxidised to S).

    If you mean did I flip the S + 2e <---> S2- equation, no it's the same as the one you posted, except I changed "e" to "2e", since S needs to gain two electrons to get a 2- charge.
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    (Original post by Mos Def)
    Sulphide ions (S^{2-}) react with conc nitric acid to form NO and S.

    They made me draw a chart of the electrode potentials of S (-0.48v), NO3(+0.8v) and N2O4(+1.03v). So obviously S reduces NO3, and NO3 reduces N2O4 right?, or does S reduce N2O4 too?

    Then it asks, the reaction takes place in 2 steps. Use electrode potential chart to calculate Ecell for each of the 2 steps. Write the overall balanced equation for the reaction.

    I can merge 2 half equations, for each one, but i can't do the part in bold.

    +REP for good explanation!

    Thanks

    erm what exam board are you on please reply
    thanks
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    Im on Nuffield children, do not worry. Men do the Nuffield.
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    Lol
 
 
 

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