1
$\begingroup$

I have conducted an experiment wherein I have used different concentrations of $\ce{H2SO4}$, $\ce{HNO3}$ and $\ce{H3PO4}$ as the electrolytes. I filled the acid in glass tube and used a 9V battery, wires and two graphite electrodes to complete the circuit. I also attached an ammeter to check the current variation during a time period of 1 minute.
I observed that the current was higher with more concentrated solutions, although my query is that when I add more of these acids, the concentration of the $\ce{OH-}$ ions reduces; it is the $\ce{OH-}$ ions and the $\ce{H+}$ ions that dissociate.
How exactly do I explain the increase in current due to increase in concentration?

$\endgroup$
3
$\begingroup$

The electrical current is carried by all ions in the solution, including the $\ce{SO_4^{2-}}$/ $\ce{HSO_4^-}$, $\ce{NO_3^-}$, and $\ce{PO_4^{3-}}$/ $\ce{HPO_4^{2-}}$/ $\ce{H_2PO_4^-}$ ions, and not just by $\ce{H+}$ and $\ce{OH-}$. So, even though the $\ce{OH-}$ concentration is decreasing with increasing acid concentration, like you say, it's much more than made up for by the increased concentration of the anions from the acids.

EDIT (in response to a comment): The contribution of ions that don't participate in the electrochemistry differs depending on the system. For example, in $\ce{NaOH}$ the $\ce{Na+}$ effectively just distributes to balance the charge from the concentration gradient in $\ce{OH-}$:

  • $\ce{H2O}$ is reduced to $\ce{H2}$ and $\ce{OH-}$ at the cathode, increasing the $\ce{OH-}$ concentration there
  • $\ce{OH-}$ is oxidized to $\ce{O2}$ and $\ce{H+}$ at the anode, decreasing the $\ce{OH-}$ concentration there
  • Thus, $[\ce{OH-}]$ changes from high at the cathode to low at the anode
  • In order for the solution to maintain electroneutrality, something has to balance the higher concentration of negative charge at the cathode, and in this system $\ce{Na+}$ is the major contributor. (Water self-ionization can't provide enough $\ce{H+}$ to maintain the charge balance.)

In other systems such as $\ce{H3PO4}$, the acid-base equilibration behavior of an electrochemically inert species can contribute to the charge carrying function:

  • Adjusted to pH 7 with, say, NaOH, phosphoric acid exists mainly as $\ce{H2PO4-}$ and $\ce{HPO4^{2-}}$
  • Water is oxidized at the anode, producing $\ce{H+}$, which reacts with $\ce{HPO4^{2-}}$ to form $\ce{H2PO4-}$
  • Water is reduced at the cathode, producing $\ce{OH-}$, which reacts with $\ce{H2PO4^{-}}$ to form $\ce{HPO4^{2-}}$ and $\ce{H2O}$
  • So, $\ce{HPO4^{2-}}$ has a higher concentration around the cathode than the anode, and $\ce{H2PO4^{-}}$ has a higher concentration around the anode than the cathode. These concentration differences result in a net transfer of charge between the two electrodes.
  • All of the negatively charged phosphate ions are balanced out by a corresponding distribution of positively-charged $\ce{Na+}$, similar to the $\ce{NaOH}$ case.
$\endgroup$
  • $\begingroup$ thank you! but say i did not use acids and used salts instead, could the same explanation be used? $\endgroup$ – nick Dec 23 '14 at 17:53
  • 1
    $\begingroup$ @nick yes. current is the rate of flow of charge. whether the charge comes from a salt or from and acid is irrelevant $\endgroup$ – bon Dec 23 '14 at 18:13
  • $\begingroup$ thank you! but at the end only particular ions go through the redox reactions at the electrodes. so how do the other ions that do not go through the redox reactions at the elctrodes contribute? $\endgroup$ – nick Dec 24 '14 at 10:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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