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- -322B- -122C- -176D- -76

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

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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

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– 322

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– 122

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– 176

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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}

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