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Gibb's Energy and Nernst Equation

The rusting o...

Question

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

A

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

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B

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

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C

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

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D

76 k J m o l^{- 1}

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Solution

The correct option is A $- 322 k J m o l^{- 1}$
$\begin{matrix} F e (s) & ⟶ & F e^{2 +} & + & 2 e^{-}; & Δ G_{1}^{\circ} 2 H^{+} & + & 2 e^{-} & + & \frac{1}{2} O_{2} & ⟶ & H_{2} O (I); & Δ G_{2}^{\circ} F e (s) & + & 2 H^{+} & + & \frac{1}{2} O_{2} & ⟶ & F e^{2 +} & + & H_{2} O & ; Δ G_{3}^{\circ} \end{matrix}$
Applying, $Δ G_{1}^{\circ} + Δ G_{2}^{\circ} = Δ G_{3}^{\circ}$
$Δ G_{3}^{\circ} = (- 2 F \times 0.44) + (- 2 F \times 1.23)$
$Δ G_{3}^{\circ} = - (2 \times 96500 \times 0.44 + 2 \times 96500 \times 1.23)$
$Δ G_{3}^{\circ} = - 322310 J$
$∴ Δ G_{3}^{\circ} = - 322 k J$

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Q. The rusting of iron takes place as follows

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

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

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

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