The half cell reactions for rusting of iron are - 2H- - 12 O2-2e- H2O - Eo - 1-23 V Fe2- -2e- - Fe - Eo-0-44 V - Go in kJ for the complete cell reaction is -

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Redox Reaction

The half cell...

Question

The half cell reactions for rusting of iron are :

$2 H^{+} + \frac{1}{2} O_{2} + 2 e^{-} \to H_{2} O; E^{o} = 1.23 V$
$F e^{2 +} + 2 e^{-} \to F e; E^{o} = - 0.44 V$

$Δ G^{o}$ (in kJ) for the complete cell reaction is :

-76.23

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-152.47

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

-228.9

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-176.4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

2 H^{+} + 1 / 2 O_{2} + 2 e^{-} \to H_{2} O; E^{o} = 1.23 V

{F e}^{2 +} + 2 e^{-} \to {F e}_{(s)}; E^{o} = - 0.44 V

Δ G^{o}

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

Δ G^{o}

The rusting of iron takes place as follows : $2 H^{+} + 2 e + \frac{1}{2} O_{2} \to H_{2} O (l); E^{o} = + 1.23 V$

$F e^{2 +} + 2 e \to F e (s); E^{o} = - 0.44 V$

Δ G^{o}

2 H^{+} + \frac{1}{2} O_{2} + 2 e^{-} \to 2 H_{2} O, E^{\circ} = + 1.23 V,

F e^{2 +} + 2 e^{-} \to F e (s); E^{\circ} = - 0.44 V .

Δ G^{\circ} (i n k J)

2 H^{+} + \frac{1}{2} O_{2} + 2 e^{-} \to H_{2} O; E^{\circ} = 1.23 V

F e^{2 +} + 2 e^{-} \to F e (s); E^{\circ} = - 0.44 V

△ G^{\circ}

Join BYJU'S Learning Program