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

The correct option is A 322 #160; kJ mol^ 1 E^⊖_cell nbsp;= 1.23 ( 0.44) = 1.67 V ⇒ nbsp;Δ G^⊖_cell = nFE^⊖_cell = ...

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Nernst Equation

The rusting o...

Question

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

- 322

k J m o l^{- 1}

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- 161

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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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^{\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}^{\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^{+} + \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^{+} + 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

Δ G^{o}

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