The half-cell reaction for the corrosion- 2H- - 12O2 - 2e- H2O- E- - 1-23 V Fe2- - 2e- Fes- E- - -0-44 VFind the G- in kJ for the overall reaction-

The correct option is A 322 kJ Fe(s)→ Fe^2+ + 2e^ ; G_1^∘ 2H^+ + 2e^ + 12O_2→ H_2O(l); G_2^∘ Fe(s) + 2H^+ + 12O_2→ Fe^2+ + H_2O; G_3^∘ ...

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Gibb's Energy and Nernst Equation

The half-cell...

Question

The half-cell reaction for the corrosion,
$2 H^{+} + \frac{1}{2} O_{2} + 2 e^{-} \to H_{2} O; E^{\circ} = 1.23 V$
$F e^{2 +} + 2 e^{-} \to F e (s); E^{\circ} = - 0.44 V$
Find the $△ G^{\circ}$ (in kJ) for the overall reaction.

- 76 k J

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- 322 k J

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- 161 k J

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- 152 k J

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Solution

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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^{+} + \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 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}^{\oplus} + 2 e^{-} + \frac{1}{2} O_{2} ⟶ H_{2} O (l); E^{⊙} = + 1.23 V

F e^{2 +} + 2 e^{-} ⟶ F e (s); E^{⊙} = - 0.44 V

Δ G^{⊙}

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

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

Δ G^{⊖}

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