The rusting of iron takes place as:2H++2e−+12O2→H2O(l); Eo=+1.23 – InterviewSolution

1.	The rusting of iron takes place as:2H++2e−+12O2→H2O(l); Eo=+1.23 VFe2++2e−→Fe(s);Eo=−0.44VThus ΔGo for the net process is
Answer» The rusting of iron takes place as: $2 H^{+} + 2 e^{-} + \frac{1}{2} O_{2} \to H_{2} O (l); E^{o} = + 1.23 V$ $F e^{2 +} + 2 e^{-} \to F e (s); E^{o} = - 0.44 V$ Thus $Δ G^{o}$ for the net process is

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