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

The correct option is A 322 kJ mol^ 1 Cathode : 2H^⊕(aq) + 2e^ + 12 O_2(g) → H_2O(l) Δ G^⊖_1 = 2 × FE^⊖ = 2 × F(1.23) Anode: ...

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Gibb's Energy and Nernst Equation

The rusting o...

Question

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

- 322 k J

m o l^{- 1}

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- 152 k J

m o l^{- 1}

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- 76 k J

m o l^{- 1}

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- 161 k J

m o l^{- 1}

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{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^{+} + 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}

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}

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}

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}

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