The half cell reactions for rusting of iron are: 2 H ++2 e-+(1/2) O 2 longrightarrow H 2 O (l) ; E°=+1.23 V Fe 2++2 e- longrightarrow Fe (s) ; E°=-0.44 V Δ G°( in kJ ) for the reaction is

Question

The half cell reactions for rusting of iron are: 2 H ++2 e-+(1/2) O 2 longrightarrow H 2 O (l) ; E°=+1.23 V Fe 2++2 e- longrightarrow Fe (s) ; E°=-0.44 V Δ G°( in kJ ) for the reaction is (A) -76 (B) -322 (C) -122 (D) -176

Tardigrade Admin · Accepted Answer

The net reaction is 2 H ++(1/2) O 2 + Fe longrightarrow H 2 O + Fe 2+ ; E°=1.67 V Δ G° =-n E° F =-(2 × 1.67 × 96500/1000) kJ =-322.31 kJ

Q. The half cell reactions for rusting of iron are :
$2 H^{+} + 2 e^{-} + \frac{1}{2} O_{2} ⟶ H_{2} O (l); E^{\circ} = + 1.23 V$
$F e^{2 +} + 2 e^{-} ⟶ F e (s); E^{\circ} = - 0.44 V$
$Δ G^{\circ} ($ in $k J)$ for the reaction is

-76

-322

-122

-176

The net reaction is
$2 H^{+} + \frac{1}{2} O_{2} + F e ⟶ H_{2} O + F e^{2 +}; E^{\circ} = 1.67 V$
$Δ G^{\circ} = - n E^{\circ} F = - \frac{2 \times 1.67 \times 96500}{1000} k J$
$= - 322.31 k J$

Q. The half cell reactions for rusting of iron are : 2H++2e−+21​O2​⟶H2​O(l);E∘=+1.23V Fe2++2e−⟶Fe(s);E∘=−0.44V ΔG∘( in kJ) for the reaction is

-76

-322

-122

-176

The net reaction is 2H++21​O2​+Fe⟶H2​O+Fe2+;E∘=1.67V ΔG∘=−nE∘F=−10002×1.67×96500​kJ =−322.31kJ

Q. The half cell reactions for rusting of iron are :
$2 H^{+} + 2 e^{-} + \frac{1}{2} O_{2} ⟶ H_{2} O (l); E^{\circ} = + 1.23 V$
$F e^{2 +} + 2 e^{-} ⟶ F e (s); E^{\circ} = - 0.44 V$
$Δ G^{\circ} ($ in $k J)$ for the reaction is

The net reaction is
$2 H^{+} + \frac{1}{2} O_{2} + F e ⟶ H_{2} O + F e^{2 +}; E^{\circ} = 1.67 V$
$Δ G^{\circ} = - n E^{\circ} F = - \frac{2 \times 1.67 \times 96500}{1000} k J$
$= - 322.31 k J$