In lead(II) bromide electrolysis, lead is attracted to the cathode and gains electrons. Why does it gain electrons when losing them would be easier since its charge is +2 meaning it would need to lose electrons to obtain an entire outer shell. This would also apply to bromine, as its charge is -1
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$\begingroup$ Less favoured operations often happen if they are forced to happen. $\endgroup$– PoutnikCommented Sep 4, 2022 at 7:05
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1 Answer
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At equilibrium of the open circuit state, there is ongoing the dynamic chemical equilibrium:
$$\ce{Pb(s) <=> Pb(aq)^2+ + 2 e-}$$
When the circuit is closed and the electrode potential is externally forced to change from the equilibrium value, then:
- If the potential is lowered, e.g. becoming the cathode during electrolysis, then the backward electrode reaction gets faster rate and the lead 8s being deposited.
- If the potential is raised, e.g. becoming the anode in the galvanic mode, or if electrolysis is reversed, then the forward electrode reaction gets faster rate and the lead is being dissolved.
