When an aqueous solution of NaBr is electrolyzed, what forms at the electrodes?

Question

can the cathode and anode be found just by knowing their reduction potential values?

Answer 1

The overall reaction equation is

2 NaBr + 2 H₂O => H₂ + Br₂ + 2 NaOH

The H₂ gas bubbles off at the cathode. The Br₂ gets created at the anode, but some will tend to stay in solution because Br₂ can go into water solution. Of course the NaOH disassociates .

Answer 2

Hi Elva V.,

To some extent, that might depend on concentrations -- very weak solutions behave like water. But usually, we are considering substantial concentrations. SO -- at the reducing electrode, Na(+) is far less reducable than is H+, so H₂ will be released (you know this, because Na(0) + H₂O = Na(+) + OH(-) + H₂ is spontaneous). But at the oxidizing electrode, Br(-) is more oxidizable than is OH(-), so Br₂ is formed instead of O₂ .

The idea of relative strengths of oxidizers and reducers is a pervasive one, and lots of commercial chemistry is based on it. For example, Cl₂ gas + a bromide-containing brine -> Br₂ + chloride is spontaneous, and a quick way to strip the bromine out (it's valuable!, and preferred for certain pool-sanitizing conditions, even if more expensive).

--Cheers, --Mr. d.