Question

In the electrolysis of aqueous sodium bromide, there are two possible anodic reactions: *

2H2O(l) ——-> 02(g) + 4H+(aq) + 4e–, E° = 1.23V

2Br–(aq) ——–> Br2(g) + 2e-2, E° = 1.08 V

Which reaction occurs at anode and why?

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Answer 1

Answer:

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:

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.

Br2−⟶Br2+2e−       º−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;

   Concentration

   Temperature

   Nature of ions