Chemistry - In NaCl electrolysis, why does chlorine ion get oxidized but not oxygen ion?
Solution 1:
I am afraid that as stated in another answer there is no simultaneous evolution of oxygen or chlorine during controlled electrolysis. It all depends on the concentration of ions in the solution. Keep in mind that the concentration of water is 55 M, so the ion concentration is nowhere near it no matter what you do!
The key reason is that the electrode potential of chlorine/chloride couple is quite different from that of oxidation of water, so if you have a high concentration of chloride ion, chlorine is oxidized first, once chlorine falls below a certain threshold, water begins to oxidize. Nernst equation can help in predicting that.
Solution 2:
I agreed with explanation given by M. Farooq. However, since OP is really a novice to the field, I'd like to clear some point for OP's benefit.
The electrolysis cell has 3 species ($\ce{Na+}$, $\ce{Cl-}$, and $\ce{H2O}$), each of which can be either oxidized or or reduced. $\ce{Na+}$ can only be reduced while $\ce{Cl-}$ can be only oxidized. Water is a unique species, which can be either oxidized or reduced based on the situation:
Relevant reduction half-reactions:
$$\ce{2 Na+ + e- <=> Na^0_{(s)}} \qquad E^\circ = \pu{-2.71 V} \tag{1}$$
$$\ce{2 H2O + 2e- <=> H2_{(g)} + 2OH- } \qquad E^\circ = \pu{-0.828 V} \tag{2}$$
Relevant oxidation half-reactions:
$$\ce{2 Cl- <=> Cl2_{(g)} + 2e-} \qquad E^\circ = \pu{-1.358 V} \tag{3}$$
$$\ce{2 H2O <=> O2_{(g)} + 4H+ + 4e-} \qquad E^\circ = \pu{-1.229 V} \tag{4}$$
Since $E^\circ$ of all four half-reactions are negative, this need external power source. That's why it is electrolytic cell. We can eliminate reduction of $\ce{Na+}$, because it need extra power to proceed ($\ce{Na+}$ can be reduced if you use molten $\ce{NaCl}$). Thus electrolytic cell would be combinations of either equations $(2)$ and $(3)$ or equations $(2)$ and $(4)$:
$$\ce{2 H2O + 2Cl- <=> H2_{(g)} + Cl2_{(g)} + 2OH- } \qquad E^\circ_ \mathrm{EMF} = \pu{-2.186 V} \tag{5}$$
$$\ce{2 H2O <=> 2H2_{(g)} + O2_{(g)} } \qquad E^\circ_ \mathrm{EMF} = \pu{-2.057 V} \tag{6}$$
The $E^\circ_\mathrm{EMF}$ value of equations $(5)$ and $(6)$ are so close that either redox reaction can happen with an electrolytic cell consists of dilute $\ce{NaCl}$ solution and a $\pu{3 V}$ battery.
Now consider the The Nernst Equation:
$$E = E^\circ - \frac{RT}{nF}\ln Q =E^\circ - \frac{2.303RT}{nF}\log_{10} Q $$
For equations $(5)$, $Q=\frac{[\ce{OH-}]^2}{[\ce{Cl-}]^2[\ce{H2O}]^2}$ and $n=2$, thus:
$$E_5 = E_5^\circ - \frac{2.303RT}{nF}\log_{10} Q = E_5^\circ - \frac{2.303RT}{2F}\log_{10} \frac{[\ce{OH-}]^2}{[\ce{Cl-}]^2[\ce{H2O}]^2}\\ = -2.186 + \frac{2.303RT}{2F}\log_{10} [\ce{OH-}]^2 [\ce{Cl-}]^2 - \frac{2.303RT}{2F}\log_{10} [\ce{H2O}]^2 \tag{7}$$
For equations $(6)$, $Q=\frac{1}{[\ce{H2O}]^2}$ and $n=4$, thus:
$$E_6 = E_6^\circ - \frac{2.303RT}{nF}\log_{10} Q = E_6^\circ - \frac{2.303RT}{4F}\log_{10} \frac{1}{[\ce{H2O}]^2}\\ = -2.057 + \frac{2.303RT}{4F}\log_{10} [\ce{H2O}]^2 \tag{8}$$
Accordingly, it is possible to increase $[\ce{Cl-}]$ such that, $E_5 \lt E_6$. As a result, redox reaction $(5)$ will predominate. As a reaction progress, $[\ce{Cl-}]$ decreases and at one point, redox reaction $(6)$ will predominate again.
Solution 3:
A technical and commercially significant point has not been cited. It directly speaks as to why, in contravention of the comparative electrode potentials (and perhaps even a Nernst concentration equation argument already presented) that chlorine is actually created over oxygen. To quote a source:
Bubble overpotential is a specific form of concentration overpotential and is due to the evolution of gas at either the anode or cathode. This reduces the effective area for current and increases the local current density. An example is the electrolysis of an aqueous sodium chloride solution—although oxygen should be produced at the anode based on its potential, bubble overpotential causes chlorine to be produced instead, which allows the easy industrial production of chlorine and sodium hydroxide by electrolysis.