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Topic: Test tube reactions  (Read 2370 times)

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Offline T

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Test tube reactions
« on: June 29, 2015, 11:13:26 PM »
Hello,
I have recently learnt that you can use electrode potentials to determine the reaction. But I don't understand how it applies in a test tube reaction. For example:

Copper will not react with sulphuric acid. The 2 equations are
Cu2+ + 2e-  ::equil:: Cu  E° = +0.34V
2H+ + 2e-  ::equil:: H2  E° = 0V

I don't understand how it will not react because won't some copper oxidise and the electrons will turn H+ into H2? The only way I can think of explaining this is that since 2H+ + 2e-  ::equil:: H2 is a stronger reductant then as soon as the copper oxidises it will be reduced again. Is this the right understanding?

Thanks

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Re: Test tube reactions
« Reply #1 on: June 30, 2015, 02:47:37 AM »
won't some copper oxidise

Nothing oxidizes by itself - you need an oxidizing agent. There is no oxidizing agent strong enough in your system.
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Offline Enthalpy

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Re: Test tube reactions
« Reply #2 on: June 30, 2015, 04:06:30 AM »
Depending on how sensitive the measure is, and how detailed the answer should be:
- The simpler and directly usable answer is "copper doesn't oxidize"
- But if considering the energy brought by heat, some copper can oxidize and quickly reduce to metal.

Trying to model how long a Cu2+ or Cu+ would survive, by estimation of shocks and so on, would be unmanageable. Fortunately, we have thermodynamics. It doesn't tell the absolute value from zero assumption but does permit many predictions from a small set of measures.

From the thermo data (H, S, G) of copper ions in water, which compare to metallic copper, you can predict numerically the concentration of dissolved copper ions at equilibrium.

A simpler approach would compare the 340mV with the temperature (300K is equivalent to 26meV) to tell that the concentration is tiny, being squeezed by a factor resembling exp(-340/26).

Offline Vidya

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Re: Test tube reactions
« Reply #3 on: June 30, 2015, 11:51:34 AM »
Hello,
I have recently learnt that you can use electrode potentials to determine the reaction. But I don't understand how it applies in a test tube reaction. For example:

Copper will not react with sulphuric acid. The 2 equations are
Cu2+ + 2e-  ::equil:: Cu  E° = +0.34V
2H+ + 2e-  ::equil:: H2  E° = 0V

I don't understand how it will not react because won't some copper oxidise and the electrons will turn H+ into H2? The only way I can think of explaining this is that since 2H+ + 2e-  ::equil:: H2 is a stronger reductant then as soon as the copper oxidises it will be reduced again. Is this the right understanding?

Thanks
Compare the reduction potential values of Cu2+ and H+ ...Do you think Cu can reduce H+ and get converted into Cu2+ which has higher reduction potential?

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