The rusting of iron takes place as-2H- -2e- -12 O2 - H2Ol- Eo-1-23 VFe2- -2e- - Fes- Eo -0-44 VThus - Go for the net process is

The given reactions are, 2H^+ +2e^ + 12 O_2 → H_2O(l); E^o = +1.23 V Fe^2+ +2e^ → Fe(s); E^o = 0.44 V For net cell reaction, E^o =E^o _OP_Fe+E^o_RP_H_2O ...

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Standard XII

Chemistry

Electrode Potential

The rusting o...

Question

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

- 322 kJ/mol

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- 161 kJ/ mol

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- 152 kJ /mol

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- 76 kJ/ mol

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Solution

The correct option is A $- 322 kJ/mol$
The given reactions are,
$2 H^{+} + 2 e^{-} + \frac{1}{2} O_{2} \to H_{2} O (l); E^{o} = + 1.23 V$
$F e^{2 +} + 2 e^{-} \to F e (s); E^{o} = - 0.44 V$
$For net cell reaction,$
$E^{o} = E_{O P_{F e}}^{o} + E_{R P_{H_{2} O}}^{o} = (0.44 + 1.23) V = 1.67 V$
$As We know,$
$Δ G^{0} = - n F E^{o}$
$∴ Δ G^{o} = - 2 \times 96500 \times 1.67 = - 322.31 kJ/mole$

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Similar questions

Q. The rusting of iron takes place as:

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

Thus,

Δ G^{o}

for the net process is:

Q. The rusting of iron takes place as follows

2 H^{+} + 2 e^{-} + \frac{1}{2} O_{2} \to H_{2} O (l); E^{\circ} = + 1.23 V

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

Calculate

Δ G^{\circ}

for the net process

Q. The half cell reactions for rusting of iron are :

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}

(in kJ) for the complete cell reaction is :

Q. The rusting of iron takes place as follows:

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

Calculate

Δ G^{⊖}

for the net process.

Q. The rusting of iron takes place as follows :

{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

Calculate

Δ G^{⊙}

for the net process.

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The rusting of iron takes place as: 2H++2e−+12O2→H2O(l); Eo=+1.23 V Fe2++2e−→Fe(s);Eo=−0.44V Thus ΔGo for the net process is

The correct option is A −322 kJ/molThe given reactions are, 2H++2e−+12O2→H2O(l); Eo=+1.23 V Fe2++2e−→Fe(s); Eo=−0.44 V For net cell reaction, Eo=EoOPFe+EoRPH2O=(0.44+1.23) V=1.67 V As We know, ΔG0=−nFEo ∴ΔGo=−2×96500×1.67= - 322.31 kJ/mole

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