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Redox Question (1 Viewer)

KFunk

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What's the difference between a metal reacting in dilute acid and a metal reacting in a concentrated acid in terms of a redox reaction?

I have this question sheet which asks for the redox reaction of copper with dil. HNO3 then with conc. HNO3

this is the difference between the reduction reactions in the answers:

dilute ---> NO3^(-) + 4H^(+) + 3e^(-) -------> NO + 2H2O

concentrated --> NO3^(-) + 2H^(+) + e^(-) -------> NO2 + H2O

What is the rule or aspect of acid reactions that accounts for this?
 

Slidey

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KFunk said:
What's the difference between a metal reacting in dilute acid and a metal reacting in a concentrated acid in terms of a redox reaction?

I have this question sheet which asks for the redox reaction of copper with dil. HNO3 then with conc. HNO3

this is the difference between the reduction reactions in the answers:

dilute ---> NO3^(-) + 4H^(+) + 3e^(-) -------> NO + 2H2O

concentrated --> NO3^(-) + 2H^(+) + e^(-) -------> NO2 + H2O

What is the rule or aspect of acid reactions that accounts for this?
The reduction reaction is between hydronium ions and NO<sub>3</sub><sup>-</sup>.

In a dilute acid, the hydronium ions would be more free, however in a concentrate acid, the acid would be more dissolved in water, so you'd have less hydronium ions available... I think.
 
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KFunk

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That makes sense...I think... but how would that translate into the NO and NO2 difference?
 

Slidey

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dilute ---> NO<sub>3</sub><sup>-</sup> + 4H<sup>+</sup> + 3e<sup>+</sup> -------> NO + 2H<sub>2</sub>O

concentrated --> NO<sub>3</sub><sup>-</sup> + 2H<sup>+</sup> + e<sup>+</sup> -------> NO<sub>2</sub> + H<sub>2</sub>O

I'm as knee-deep in confusion as you are. But this seems to be similar to the stuff about acid rain.
 

KFunk

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How'd you get the nice, neat looking '+' and '-' signs up top?
 

Slidey

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Subscript and superscript.

{sub}hello{/sub} with HTML tags instead of parenthesis produces: <sub>hello</sub>
 

KFunk

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I will work this out... SO<sub>4</sub><sup>2-</sup>. Wow, quite handy, danke.

y = a<sub>n</sub>x<sup>n</sup> + a<sub>n-1</sub>x<sup>n-1</sup> ... just checking a forum {pun}function{/pun}.
 
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wow, i never realised a more concetrated acid would change the output NO2
do they really tos these questions out?
 

Slidey

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A different acid concentration WOULD probably change how the redox reaction occurs, however I don't think they can ask this in a test because 1) we don't cover it in the syllabus and 2) I don't think there's any generic rule for figuring it out.

I have other gripes with this question, but...
 
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my chemistry teacher probably would launch these questions out, the tests she sets are VERY hard
 

KFunk

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Slide Rule said:
A different acid concentration WOULD probably change how the redox reaction occurs, however I don't think they can ask this in a test because 1) we don't cover it in the syllabus and 2) I don't think there's any generic rule for figuring it out.

I have other gripes with this question, but...
In which case it was a mean quesiton. What are your other gripes?
 

xiao1985

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hm... interestin q... imo:

this question could be simply the geometry of the reaction... every reaction need to have the correct molecules colliding at the right way in the right proportion... in dilute acid, since the H +'s and NO3 -'s are rather spread out, u probably have only a few hydromium ions, no3 - for a set area of metal surface... hence u will have a proficiency in electrons, which favours the first reaction...

on the other hand, concentrated acid will make u have more no3 - and h + per unit area of cu surface... hence there will be a deficiency of electrons, which favours the 2nd eq...
 

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