The half-cell reactions for the corrossion are 2H+ + 1/2 O2 + 2e- → H2O ; E° = - 1.23 V Fe2+ + 2e- → Fe; E° = - 0.44 V Find the Δ G° (in kJ) for the overall reaction.

Question

The half-cell reactions for the corrossion are 2H+ + 1/2 O2 + 2e- → H2O ; E° = - 1.23 V Fe2+ + 2e- → Fe; E° = - 0.44 V Find the Δ G° (in kJ) for the overall reaction. (A)  - 76 (B) -322 (C)  -161 (D) -152

Tardigrade Admin · Accepted Answer

The reduction potential of two half reactions suggest that reduction of

H^{+}

and oxidation of

F e

take place, so

E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}

= 1.23 - (- 0.44) = 1.67 V

Δ G^{\circ} = - n F E^{\circ} n = 2

Δ G^{\circ} = - 2 \times 96500 \times 1.67 = - 322310 J

= - 322.31 k J

Q. The half-cell reactions for the corrossion are
$2 H^{+} + 1/2 O_{2} + 2 e^{-} \to H^{2} O; E^{\circ} = - 1.23 V$
$F e^{2 +} + 2 e^{-} \to F e; E^{\circ} = - 0.44 V$
Find the $Δ G^{\circ}$ (in $k J$ ) for the overall reaction.

$- 76$

$- 322$

$- 161$

$- 152$

Q. The half-cell reactions for the corrossion are 2H++1/2O2​+2e−→H2O;E∘=−1.23V Fe2++2e−→Fe;E∘=−0.44V Find the ΔG∘ (in kJ) for the overall reaction.

−76

−322

−161

−152

The reduction potential of two half reactions suggest that reduction of H+ and oxidation of Fe take place, so Ecell∘​=Ecathode∘​−Eanode∘​ =1.23−(−0.44)=1.67V ΔG∘=−nFE∘n=2 ΔG∘=−2×96500×1.67=−322310J =−322.31kJ

Q. The half-cell reactions for the corrossion are
$2 H^{+} + 1/2 O_{2} + 2 e^{-} \to H^{2} O; E^{\circ} = - 1.23 V$
$F e^{2 +} + 2 e^{-} \to F e; E^{\circ} = - 0.44 V$
Find the $Δ G^{\circ}$ (in $k J$ ) for the overall reaction.

$- 76$

$- 322$

$- 161$

$- 152$