The half cell reaction for rusting of iron are 2 H++(1/2)O2+2e- arrow H2O;E° =1.23 V Fe2 ++2e- arrow Fe;E° =-0.44 V ÎG° for the reaction 4H++O2+2Fe arrow 2Fe2 ++2H2O is :-

Question

The half cell reaction for rusting of iron are 2 H++(1/2)O2+2e- arrow H2O;E° =1.23 V Fe2 ++2e- arrow Fe;E° =-0.44 V ÎG° for the reaction 4H++O2+2Fe arrow 2Fe2 ++2H2O is :- (A) -76 kJ (B) -644 kJ (C) -122 kJ (D) -176 kJ

Tardigrade Admin · Accepted Answer

4H++O2+4e- arrow 2H2O 1.23 (2 Fe arrow 2 Fe2 + + 4 e-/4 H+ + O2 + 2 Fe arrow 2 Fe2 + + 2 H2 O)(0.44/E° =1.67) ÎG° =-nFE° =-4× 96500× 1.67 =-644 kJ

Q. The half cell reaction for rusting of iron are
$2 H^{+} + \frac{1}{2} O_{2} + 2 e^{-} \to H_{2} O; E^{\circ} = 1.23 V$
$F e^{2 +} + 2 e^{-} \to F e; E^{\circ} = - 0.44 V$
$Δ G^{\circ}$ for the reaction
$4 H^{+} + O_{2} + 2 F e \to 2 F e^{2 +} + 2 H_{2} O$ is :-

$- 76 k J$

$- 644 k J$

$- 122 k J$

$- 176 k J$

$4 H^{+} + O_{2} + 4 e^{-} \to 2 H_{2} O$ $1.23$
$\frac{2 F e \to 2 F e ^{2 +} + 4 e ^{-}}{4 H ^{+} + O _{2} + 2 F e \to 2 F e ^{2 +} + 2 H _{2} O} \frac{0.44}{E ^{\circ} = 1.67}$
$Δ G^{\circ} = - n F E^{\circ} = - 4 \times 96500 \times 1.67$
$= - 644 k J$

Q. The half cell reaction for rusting of iron are 2H++21​O2​+2e−→H2​O;E∘=1.23V Fe2++2e−→Fe;E∘=−0.44V ΔG∘ for the reaction 4H++O2​+2Fe→2Fe2++2H2​O is :-

−76kJ

−644kJ

−122kJ

−176kJ