The half cell reactions for the corrosion are 2H-1-2O2-2e-H2O-Eo-1-23V Fe2-2e-Fes-Eo-0-44V- Find teh -Go in kJ for the overall reaction-

The correct option is D 152 Δ G ^ 0 = nF E _ Cell ^ 0 = 2× 96487C mol ^ 1 × 0.79V = 152449.46J mol ^ 1 ...

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Ion Electron Method

The half cell...

Question

The half cell reactions for the corrosion are
$2 H^{+} + 1 / 2 O_{2} + 2 e^{-} \to H_{2} O; E^{o} = 1.23 V$
${F e}^{2 +} + 2 e^{-} \to {F e}_{(s)}; E^{o} = - 0.44 V$ . Find teh $Δ G^{o}$ (in kJ) for the overall reaction:

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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 H_{2} O; E^{\circ} = 1.23 V

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

△ 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}

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)

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