The rusting of iron takes place as follows 2 H -2 e -1-2 O 2- H 2 O l - E -1-23 V F e 2-2 e - Fe s - E -0-44 VCalculate - G - for the net process A- -152 kJ mol -1B- -322 kJ mol -1C- -161 kJ mol -1D- -76 kJ mol -1

The correct option is B 322 kJ mol 1[ amp; amp; amp; amp; amp; Fe(s) amp; ⟶ amp; Fe^2+ amp;+ ...

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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

-322 kJ mol^-1

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-161 kJ mol^-1

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-152 kJ mol^-1

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-76 kJ mol^-1

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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}^{\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^{⊙}

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}

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