Redox equation help ?? Watch

coconut64
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Gold metal in the ore reacts with cyanide ions, water and oxygen to form a water-soluble complex ion, [Au(CN)2],

Construct the overall equation for this reaction.

My answer : 3e- + Au + 2CN- + O2 +2H2O -> [Au(CN)2] + 4OH
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I am not sure about the bit in brown. Thanks
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Pigster
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You needed to multiply the Au half equation by 4 so that you'll be able to cancel out the e-.
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Jassy16
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i think when its asking for overall equation it does not want any unaccounted electrons, so only then mentioned reactants and products are supposed to be in the equation.

Bearing that in mind i got your equation ( 3e- + Au + 2CN- + O2 +2H2O -> [Au(CN)2] + 4OH-)
and instead of balancing with electrons i balanced the charge by slowly increasing the amount of CN- ions. By doing this you eventually realise its 8CN- ions you need on the left to ensure the charge is balanced on both sides.

The only reason i think this is what they want you to do is because the question states overall equation which means they want no electrons in the equation.
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dreamfyre
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is this aqa? and did it say in the question alkali conditions? im confused on how u got oh-
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coconut64
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(Original post by Jassy16)
i think when its asking for overall equation it does not want any unaccounted electrons, so only then mentioned reactants and products are supposed to be in the equation.

Bearing that in mind i got your equation ( 3e- + Au + 2CN- + O2 +2H2O -> [Au(CN)2] + 4OH-)
and instead of balancing with electrons i balanced the charge by slowly increasing the amount of CN- ions. By doing this you eventually realise its 8CN- ions you need on the left to ensure the charge is balanced on both sides.

The only reason i think this is what they want you to do is because the question states overall equation which means they want no electrons in the equation.
How do you know that O2 and H2O will also be in the half equation of Au?
Thanks
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Jassy16
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(Original post by coconut64)
How do you know that O2 and H2O will also be in the half equation of Au?
Thanks
water and oxygen are not in the half equation for Au (are they?).

the way i done the question was by writing out all the reactants and products then balancing.

Admittedly i was stuck so i worked back from the answer and the only way i could justify it was is by using additional CN- instead of electrons to balance the charge of the reaction.

I cant see why half equations would be necessary for this question though. Because neither of the half equations for CN or Au would have water or oxygen in then.

The water and oxygen only become a factor because the specifically mention them as reactants.
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coconut64
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(Original post by Jassy16)
water and oxygen are not in the half equation for Au (are they?).

the way i done the question was by writing out all the reactants and products then balancing.

Admittedly i was stuck so i worked back from the answer and the only way i could justify it was is by using additional CN- instead of electrons to balance the charge of the reaction.

I cant see why half equations would be necessary for this question though. Because neither of the half equations for CN or Au would have water or oxygen in then.

The water and oxygen only become a factor because the specifically mention them as reactants.
Oh okay thanks anyway!
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coconut64
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(Original post by Pigster)
You needed to multiply the Au half equation by 4 so that you'll be able to cancel out the e-.
Just wondering how do you know that you have to write the half equations out separating as normally for the overall equation, don't you just write everything out all at once. Also, how do you know water & O2 are in the half equation of Au?


Thanks
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Pigster
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(Original post by coconut64)
Just wondering how do you know that you have to write the half equations out separating as normally for the overall equation, don't you just write everything out all at once.
You don't have to write the half equations, you can just do it as an overall balanced redox equation. Assuming you can. Most people (including me) find it much easier to create the two half equations, multiply both to allow the e- to balance, and combine. Generally, whenever my students try to do it all in one go they end up (as you did) with either e- left over or charges that don't balance.

Also, how do you know water & O2 are in the half equation of Au?
O2 is not in the half equation, only the species having its oxidation state changed, water and H+ (or OH-) go into half equations.
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coconut64
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(Original post by Pigster)
You needed to multiply the Au half equation by 4 so that you'll be able to cancel out the e-.
3e- + Au + 2CN- + O2 +2H2O -> [Au(CN)2] + 4OH

I see what you mean now but when how do I know I need change the no of CN- and [Au(CN)2]– instead the of the no of OH-?

Thanks
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Pigster
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(Original post by coconut64)
3e- + Au + 2CN- + O2 +2H2O -> [Au(CN)2] + 4OH

I see what you mean now but when how do I know I need change the no of CN- and [Au(CN)2]– instead the of the no of OH-?

Thanks
All the information you need is in the Q: it tells you what the reactants are and what the products are.
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