-1
$\begingroup$

Why does the hydrogen gets reduced at the cathode and not $\ce{Na}$ when aqueous solution of brine is subjected to electrolysis?

My book says because discharge potential of hydrogen is lower than that of sodium but in the reduction potential table, hydrogen is assigned a value = 0 and sodium a value of -2.7. So isn't the negative value of sodium lesser than that of hydrogen? Or is it being talked just magnitude wise?

Please explain me the whole concept of what factors decide the reduction and oxidation of some elements over others.

$\endgroup$
  • $\begingroup$ The whole concept of what factors decide how a redox happens is a bit too broad. $\endgroup$ – M.A.R. May 22 '16 at 9:52
0
$\begingroup$

A positive value for the reduction potential means that it has a higher propensity for reduction, whereas a lower value for the reduction potential means that it is less likely to be reduced (i.e. it will resist reduction).

Since $\ce{Na+}$ has a lower reduction potential than $\ce{H+}$, we can expect that $\ce{H+}$ will be preferentially reduced.

For other elements (especially those with more than one oxidation state), it's a little more complicated. You would need to know the condition under which the electrolysis is done. Familiarity with Latimer diagrams will also help you figure out which is the most likely oxidation state.

$\endgroup$
  • $\begingroup$ Okay so why does my book say "since the discharge potential of H+ ions is lower than that of Na+ ions, therefore H+ ions are discharged in preferance to Na+ ions" what does it mean? And could you please explain me the latimer diagrams with an example if possible? $\endgroup$ – Ujjwal May 22 '16 at 5:26
  • $\begingroup$ If "discharge potential" refers to the discharge of positive ions, then the more negative values on the Reduction Potential Tables means higher discharge potential, and the more negative values mean lower discharge potential. Also, I think for simple systems you won't need the Latimer diagram and all you have to do is to compare the relative potentials according to the principle I've described in the answer. $\endgroup$ – IT Tsoi May 22 '16 at 5:33
  • $\begingroup$ I think there is some mistake in the comment you made $\endgroup$ – Ujjwal May 22 '16 at 6:17

Not the answer you're looking for? Browse other questions tagged or ask your own question.