# Why does aqueous NaCl conduct electricity?

Could someone please explain why an aqueous solution of NaCl conducts electricity? I've Googled this up but couldn't get a satisfactory answer. This has an answer on Physics.SE but that's too complicated. My textbook doesn't answer my question.

• Essentially you understand that it is because it dissociates, and your follow-up question is why it dissociates, which, as someone else pointed out, is another question. – DHMO Oct 8 '16 at 11:10
• See the answer, and linked reference therein, by @M.Farooq: chemistry.stackexchange.com/a/118441/79678. – Ed V Jul 15 '20 at 14:34

Original Question:
Why an aqueous solution of $$\ce{NaCl}$$ conducts electricity

Because $$\ce{NaCl}$$ is an electrolyte. This means it yields ions in solution.

Simply put, solid $$\ce{NaCl}$$ consists of $$\ce{Na+}$$ cations and $$\ce{Cl-}$$ anions bound together in a rigid crystal lattice. When it melts or is dissolved in water, the crystal lattice breaks. The ions are now able to move around. Similar to charged particles in a metal conductor (in this case electrons), in liquid form or aqueous solution the ions are the charged particles that can move, allowing the solution to conduct electricity.

Follow-up question:
Why $$\ce{NaCl}$$ dissociates in water

The (rather oversimplified) answer is that dissociation of an ionic compound is facilitated by ion-dipole attractions between the ions of the compound and the polar water molecules.

The water molecules (dipoles) are attracted to the ions and cause the crystal lattice to destabilize and ions to dissociate. The charged ions in solution are surrounded and stabilized by the water molecules (dipoles).

Note: Some ions migrate as ion pairs, but for a strong electrolyte most or all the ions will be dissociated and surrounded by water molecules.

As you seem dissatisfied with the answers received so far, I assume you want to fully understand the mechanism involved in dissociation. Unfortunately, this is rather complex.

There is an excellent paper by Ballard & Dellago [1] that explains their work on the subject, but you will probably need a bit more physical chemistry / thermodynamics knowledge to understand it fully.

Reference

1. Ballard, AJ & Dellago, C "Toward the Mechanism of Ionic Dissociation in Water," *J. Phys. Chem. *B 2012, 116, 45, 13490–13497 Publication Date:October 19, 2012 https://doi.org/10.1021/jp309300b

NaCl is an electrolyte. When in solution it dissociates into Na+ and Cl-. When you put electrodes in the solution, the cations are drawn to the cathode and the anions to the anode. This movement produces a current and that is why NaCl solutions can conduct electricity.

• Why does it dissociate? – user29731 Oct 7 '16 at 19:29
• I think this answer needs some elaboration. Does this only work with AC? If not, what happens with DC when the bulk of the ions have already migrated? – bpedit Oct 7 '16 at 19:34
• With a DC current, you'll probably get some electrolysis of water and possible oxidation of the chloride (which helps by removing negative charge from the positive electrode). – Zhe Oct 7 '16 at 20:58
• @Zhe Agreed. I was just trying to prompt the psoter into fleshing out his answer for the OP. – bpedit Oct 7 '16 at 21:21

$$\ce{NaCl}$$ dissociates into $$\ce{Na+}$$ and $$\ce{Cl-}$$ when it dissolves into water. Hydration stabilizes the ions formed. There is $$\delta+$$ charge on hydrogens of water and $$\delta-$$ charge on the atoms of oxygen. When $$\ce{NaCl}$$ is dissolved into water, the hydrogens are attracted to the $$\ce{Cl-}$$ ions and oxygen atoms to the $$\ce{Na+}$$ ions. Thus water molecules surround and separate the ions.

Hence, in an aqueous solution of $$\ce{NaCl}$$, there are positive and negative ions available to move freely (of course with some resistance due to other molecules in vicinity). When a potential difference is applied, the positive ions are attracted to the negative terminal and vice versa. Thus, a current is established.

• @Abcd See this video on YouTube. link – Apoorv Potnis Oct 8 '16 at 15:31
• I do not think that actual $\ce{OH-}$ ions are attracted to $\ce{Na+}$ because the dissociation constant of water is quite low, $pK_w = 14$. So, I don't think that free $H+$ and $OH-$ ions are available. A water molecule as a whole orients itself in a particular way and surrounds the ions. – Apoorv Potnis Oct 8 '16 at 15:49
• Sorry, I didn't mean that $OH-$ ions are not attracted to $Na+$ but I think that the effect wouldn't be as much. – Apoorv Potnis Oct 8 '16 at 16:14
• Thankyou. So can we say that dissociation of NaCl takes place because of orientation of water molecules? – user29731 Oct 8 '16 at 16:16
• The simple answer is that water molecules have a dipole. The hydrogen atoms are slightly positively charged and the oxygen atoms slightly negatively charged. So, as you can see in the diagram above, water molecules around positively charged sodium orient themselves such that the oxygen is closer to sodium than the hydrogens. This is known as an ion-dipole interaction. The opposite is true for chlorine. @Abcd – orthocresol Oct 8 '16 at 16:23

$$\ce{NaCl}$$ is strongly ionic compound. It gets completely ionized and dissociates in $$\ce{Na+}$$ and $$\ce{Cl-}$$. Both $$\ce{Na+}$$ and $$\ce{Cl-}$$ are surrounded by water molecules, $$\ce{Na+}$$ is surrounded with $$\ce{O}$$ of $$\ce{H2O}$$ facing towards $$\ce{Na+}$$, similarly $$\ce{Cl-}$$ is surrounded by $$\ce{H2O}$$ molecules with $$\ce{H}$$ facing towards $$\ce{Cl-}$$. Presence of positive and negative ions helps in conduction of electricity.

When electrodes are inserted and current is passed through these electrodes, movement of ions in opposite direction creates current.

After passing current for a long time (especially DC current), the concentration of $$\ce{Cl-}$$ ions will decrease. The concentration of $$\ce{OH-}$$ will increase gradually and there will be competition between $$\ce{Cl-}$$ and $$\ce{OH-}$$.

Electrolysis of $$\ce{NaCl}$$

Anode: $$\ce{2 Cl- -> Cl2 + 2 e-}\quad E=-1.36\ \mathrm V$$

Cathode: $$\ce{H2O + 2 e- -> H2 + 2 OH-}\quad E=-0.83\ \mathrm V$$

Electrolysis of $$\ce{NaOH}$$

Anode: $$\ce{4 OH- -> O2 + 2 H2O + 2 e-}\quad E=-0.40\ \mathrm V$$

Cathode: $$\ce{H2O + 2 e- -> H2 + 2 OH-}\quad E=-0.83\ \mathrm V$$

After Sufficient long time there will be competition between $$\ce{Cl-}$$ and $$\ce{OH-}$$ to get oxidized on Anode, whereas there is only one possible reaction on cathode. Reaction on Anode depends upon the concentration and reduction potential of $$\ce{Cl-}$$ and $$\ce{OH-}$$. Actually on cathode there is also possibility of reduction of $$\ce{Na+}$$ but reduction potential is $$-2.7\ \mathrm V$$ which is difficult than $$\ce{H2O}$$ so only one reaction takes place on cathode.