The rusting of iron takes place as follows - 2H -2e-12O2- H2Ol- E -1-23 V Fe2-2e- Fes- E -0-44 VCalculate - G - for the net process-

The correct option is C 322 kJ mol^ 1 Fe(s)⟶ Fe ^ 2+ +2 e ^ ;Δ G _ 1 ^ 0 2 H ^ + + 1 / 2 O _ 2 ⟶ H _ 2 O(l) ;Δ G _ 2 ^ 0 Fe(s)+2 H ...

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Gibb's Energy and Nernst Equation

The rusting o...

Question

The rusting of iron takes place as follows :
${2 H}^{\oplus} + 2 e^{-} + \frac{1}{2} O_{2} ⟶ H_{2} O (l); E^{⊙} = + 1.23 V$
$F e^{2 +} + 2 e^{-} ⟶ F e (s); E^{⊙} = - 0.44 V$
Calculate $Δ G^{⊙}$ for the net process.

- 322 k J m o l^{- 1}

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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^{\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^{⊖}

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^{+} + 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}

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