chemistry

Balance the following redox equation occurring in a basic solution:

AG + CN{-} + O2 = AG(CN)2{-}

I tried...but this is wrong...2H2O+ AG + CN{-} + O2 = AG(CN)2{-}+4OH{-}... I am supposed to do half reactions...how do I spit the product AG(CN)2{-}?????

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  1. Ag half cell is
    Ag ==> Ag^+ + e
    O2 half cell is
    Stepwise. Copy this and memorize how to do it.
    1. Identify the element changing oxidation state. I'll make it easy and do it for you. O2 is zero and it goes to OH^-.

    2. Preliminarily balance the element changing.
    O2 ==> 2OH^- (so we are comparing the same number of atoms).

    3. Add electrons to the appropriate side to balance the change in oxdn state. O2 is zero and oxygen in 2OH^- is -4. Therefore, add electrons to the left to make the electron change balanced.
    O2 + 4e ==> 2OH^-

    4. Count the charge on each side and add to the appropriate side
    a)H^+ if it is an acid solution or
    b)OH^- if it is a basic solution.
    I count -4 on the left and -2 on the right. This is a basic solution; therefore, I add 2 OH^- to the right to balance the charge.
    O2 + 4e ==> 2OH^- + 2OH^-

    5. Then add H2O (usually to the opposite side but not always) to balance the H atoms.
    2H2O + O2 + 4e ==> 4OH^-

    6. Check
    a. oxidation change
    b. atoms on left = atoms on right.
    c. charge on left = charge on right.
    If a,b, and c check, the half equation is correct.

    So now you multiply the Ag half cell by 4 to make electrons lost = electrons gained, add the CN^- needed on the left to balance those in the product on the right and your equation is complete.

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  2. I think I understand...but how do I do CN- reactant? do I do a half reaction too?-->with a product of CN2???

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  3. the answer is 4AG + 8CN{-} + O2 + 2H2O=> 4AG(CN)2{-} + 4OH{-}

    I do not understand the plugging in of CN. COuld You please explain?

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  4. By the way, I don't like to see you write AG for Ag. AG is not the same as Ag.
    Multiply the Ag half by 4, the O2 half cell by 1 and add them. That gives
    4Ag + 2H2O + 4e + O2 ==> 4Ag^+ + 4OH^- 4e.
    The 4e cancel since they are on both sides (that's why we wanted electrons lost = electrons gained) and the equation now is
    4Ag + 2H2O + O2 ==> 4OH^- + 4Ag^+
    The CN^- does not undergo any oxdn or redn so I balanced everything without CH^- because that can be added when we finish. How to do that? I look at the product side and see the Ag^+ present as Ag(CN)2^- and I have a coefficient of 4 which means there are 8 CN^- on the right side. So I add 8CN^- to the left side and things should be set.
    4Ag + 2H2O + O2 + 8CN^- ==> 4OH^- + 4Ag(CN)2^-
    I always check it one more time when I finish.
    Atoms: 4Ag on left and right.
    4H atoms on left and right.
    4O atoms on left and right.
    8CN^- on left and right.
    charge:
    8- on left and 8- on right.
    electron change: Ag gained 4, O (of O2) lost 4 in going to 2OH^-
    Everything balances.
    Check my work. Make sure it's correct.

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  5. line 3(first equation) I wrote as
    4Ag + 2H2O + 4e + O2 ==> 4Ag^+ + 4OH^- 4e.
    and it should be +4e on the right side. I omitted the + sign

    line 7.
    The CN^- does not undergo any oxdn or redn so I balanced everything without CH^- because that can be added when we finish. How to do that? I look at the product side and see the Ag^+ present as Ag(CN)2^- and I have a coefficient of 4 which means there are 8 CN^- on the right side. So I add 8CN^- to the left side and things should be set.
    .....so I balance everything without CN^-..........

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  6. Great! The CN was the spectator that did "nothing." thank you very much.

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  7. I'm glad to have helped.

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