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Why is bromine produced at the anode when aqueous sodium bromide undergoes electrolysis?
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Kevin Wright
Why is bromine produced at the anode when aqueous sodium bromide undergoes electrolysis?
This is how I see it:
In the electrolysis of NaBr, water is reduced at the cathode. This occurs because water is more easily reduced than are sodium ions. This is reflected in their standard reduction potential.
At cathode reduction of water occurs:
$$\ce{2H2O(l) + 2e- -> H2(g) + 2OH-(aq)}$$
And hydrogen gas is produced
At the anode, where oxidation occurs, the standard oxidation potential of water is –1.23 volts, while that for bromide ions is -1.07 volts.
The production of bromine itself has a negative electrode potential.
One of the half-reactions must be reversed to yield an oxidation.
$$\ce{Br^2- -> Br2 + 2e-}~~~~~~~ Eº -1.07$$
Remember that when one reverses a reaction, the sign of Eº (+ or –) for that reaction is also reversed.
This means that bromide ions are more easily oxidized than water.
It is important to note: When current begins to flow, the distribution of ions around the electrodes changes, and the equilibrium electrode potentials no longer accurately apply.
In the electrolysis of NaBr, water is reduced at the cathode. This occurs because water is more easily reduced than are sodium ions. This is reflected in their standard reduction potential.
At cathode reduction of water occurs:
$$\ce{2H2O(l) + 2e- -> H2(g) + 2OH-(aq)}$$
And hydrogen gas is produced
At the anode, where oxidation occurs, the standard oxidation potential of water is –1.23 volts, while that for bromide ions is -1.07 volts.
The production of bromine itself has a negative electrode potential.One of the half-reactions must be reversed to yield an oxidation.
$$\ce{Br^2- -> Br2 + 2e-}~~~~~~~ Eº -1.07$$
Remember that when one reverses a reaction, the sign of Eº (+ or –) for that reaction is also reversed.
This means that bromide ions are more easily oxidized than water.
It is important to note: When current begins to flow, the distribution of ions around the electrodes changes, and the equilibrium electrode potentials no longer accurately apply.
The electrode potential for the reduction of water is -0.83V whereas for H+ ions, it is 0.00V. As such, shouldnt Hydrogen ions be reduced? Also, why are you considering the +1.23V reaction of water rather than the +0.40V one?More
Why is water reduced at the cathode? Shouldnt it be Hydrogen ions since they have an electrode potential of 0 V? Also, could you write the reaction in which water is reduced? The 1.23V reaction I know has Oxygen being reduced, not water.More
And where are you generating the hydrogen ions from in the first place? Please show the reaction that you are referring to. Also read Ivan Neretins comments they essentially answer your additional questionsMore
This is how I see it:
In the electrolysis of NaBr, water is reduced at the cathode. This occurs because water is more easily reduced than are sodium ions. This is reflected in their standard reduction potential.
At cathode reduction of water occurs:
$$\ce{2H2O(l) + 2e- -> H2(g) + 2OH-(aq)}$$
And hydrogen gas is produced
At the anode, where oxidation occurs, the standard oxidation potential of water is –1.23 volts, while that for bromide ions is -1.07 volts.
The production of bromine itself has a negative electrode potential. One of the half-reactions must be reversed to yield an oxidation.
$$\ce{Br^2- -> Br2 + 2e-}~~~~~~~ Eº -1.07$$
Remember that when one reverses a reaction, the sign of Eº (+ or –) for that reaction is also reversed.
This means that bromide ions are more easily oxidized than water.
It is important to note: When current begins to flow, the distribution of ions around the electrodes changes, and the equilibrium electrode potentials no longer accurately apply.
I think other factors also apply;
This is how I see it:
In the electrolysis of NaBr, water is reduced at the cathode. This occurs because water is more easily reduced than are sodium ions. This is reflected in their standard reduction potential.
At cathode reduction of water occurs:
$$\ce{2H2O(l) + 2e- -> H2(g) + 2OH-(aq)}$$
And hydrogen gas is produced
At the anode, where oxidation occurs, the standard oxidation potential of water is –1.23 volts, while that for bromide ions is -1.07 volts.
The production of bromine itself has a negative electrode potential.One of the half-reactions must be reversed to yield an oxidation.
$$\ce{Br^2- -> Br2 + 2e-}~~~~~~~ Eº -1.07$$
Remember that when one reverses a reaction, the sign of Eº (+ or –) for that reaction is also reversed.
This means that bromide ions are more easily oxidized than water.
It is important to note: When current begins to flow, the distribution of ions around the electrodes changes, and the equilibrium electrode potentials no longer accurately apply.
I think other factors also apply;
More
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