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November 01, 2025, 11:10:08 am

Author Topic: Redox HELP!!! thanks you  (Read 5526 times)  Share 

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jack_chay

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Redox HELP!!! thanks you
« on: September 10, 2013, 09:51:27 pm »
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Ba(NO3)2(aq) + H2SO4(aq) → BaSO4(s) + 2HNO3(aq)

Is barium nitrate oxidised, reduced, neither or both?

thanks in advance :)

psyxwar

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Re: Redox HELP!!! thanks you
« Reply #1 on: September 10, 2013, 09:55:28 pm »
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Well, there's no change in oxidation number for any of the elements in Ba(NO3)2 right? So neither.
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jack_chay

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Re: Redox HELP!!! thanks you
« Reply #2 on: September 10, 2013, 10:02:45 pm »
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oh ok, thank you :)

jack_chay

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Re: Redox HELP!!! thanks you
« Reply #3 on: September 10, 2013, 10:05:51 pm »
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I have another redox question:

how would you find conjugate oxidant and reductant (redox pairs)

so for example:

a( the conjugate pair when Fe2+ is the oxidant
b( the conjugate pair when Fe2+ is the reductant

thank you :)

lzxnl

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Re: Redox HELP!!! thanks you
« Reply #4 on: September 10, 2013, 10:09:31 pm »
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Ba(NO3)2(aq) + H2SO4(aq) → BaSO4(s) + 2HNO3(aq)

Is barium nitrate oxidised, reduced, neither or both?

thanks in advance :)

This is a precipitation reaction, where the barium ions react with sulfate ions to form barium sulfate. No reduction or oxidation occurs; the ions are just bonded differently.

As for your second question, if Fe(II) is the oxidant, it is reduced, so the conjugate pair is Fe metal
If it is the reductant, it is oxidised, so the conjugate pair is Fe(III)

Note: oxidants CAUSE oxidation, are themselves reduced
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jack_chay

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Re: Redox HELP!!! thanks you
« Reply #5 on: September 10, 2013, 10:12:31 pm »
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um.. just wondering so is the barium question a precipitation reaction, not a redox at all?

lzxnl

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Re: Redox HELP!!! thanks you
« Reply #6 on: September 10, 2013, 10:15:38 pm »
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Yes. You have bariums ions + sulfate ions => barium sulfate solid.
It's a precipitation reaction; the nitrate and the hydrogen ions don't participate.
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jack_chay

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Re: Redox HELP!!! thanks you
« Reply #7 on: September 11, 2013, 07:48:02 am »
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yup i get it now :) thank you

jack_chay

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Re: Redox HELP!!! thanks you
« Reply #8 on: September 11, 2013, 07:50:38 am »
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MnO4-  +  8H  +  5Fe2+  --->    Mn2+  +  4H2O  +  5Fe3+

How would i work out which is the oxidised and which is reduced, because there are three reactants?

Thank you

Good Will

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Re: Redox HELP!!! thanks you
« Reply #9 on: September 11, 2013, 08:04:57 am »
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If you assign oxidation numbers, we can see that the Fe had an oxidation number of 2+ which changed to a 3+. This means that the Fe has been oxidised and consequently, the MnO4- has been reduced.
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jack_chay

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Re: Redox HELP!!! thanks you
« Reply #10 on: September 11, 2013, 11:07:55 am »
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what about the h?

psyxwar

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Re: Redox HELP!!! thanks you
« Reply #11 on: September 11, 2013, 11:19:00 am »
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what about the h?
Oxidation number is still one. It goes from H+ to H2O.
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jack_chay

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Re: Redox HELP!!! thanks you
« Reply #12 on: September 11, 2013, 07:25:34 pm »
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oh ok thank you everyone :)

jack_chay

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Re: Redox HELP!!! thanks you
« Reply #13 on: September 11, 2013, 07:30:11 pm »
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What about this equation:

Cd + NiO2 +2H2O ---> Cd(OH)2 + Ni(OH)2

I know Cd is oxidised, Ni doesn't change so H2O is reduced?

psyxwar

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Re: Redox HELP!!! thanks you
« Reply #14 on: September 11, 2013, 07:34:27 pm »
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Nickel does change. It goes from +IV to +II (reduction)
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