The half-cell reactions for rusting of iron are- 2 H -1-2 O 2-2 e - 2 H 2 O - E -1-23 VF e 2-2 e - Fe s - E -0-44 V- G - in kJ for the reaction is -A- -76B- -322C- -122D- -176

The correct option is B – 3222H^+ +12O_2 + 2e^ amp;→ H_2OE^∘=1.23 V Fe^2+ + 2e^ amp;→ Fe(s)E^∘= 0.44 V (Subtracting) Fe(s)+2H^+ ...

You visited us 59 times! Enjoying our articles? Unlock Full Access!

Gibb's Energy and Nernst Equation

The half-cell...

Question

The half-cell reactions for rusting of iron are,
$2 H^{+} + \frac{1}{2} O_{2} + 2 e^{-} \to 2 H_{2} O, E^{\circ} = + 1.23 V,$
$F e^{2 +} + 2 e^{-} \to F e (s); E^{\circ} = - 0.44 V .$
$Δ G^{\circ} (i n k J)$ for the reaction is :

− 76

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

– 322

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

– 122

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

– 176

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

Open in App

Solution

Suggest Corrections

Similar questions

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^{o} = 1.23 V

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

Δ G^{o}

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}

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

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

Δ G^{\circ}

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}

Join BYJU'S Learning Program