The rusting of iron takes place as- 2H- - 2e- - 1-2 O2 - H2O l - Eo - - 1-23 V Fe2- - 2e- - Fes - Eo - - 0-44 VThus- - Go for the net process is-

The correct option is A 322 kJ/mol 2H^+ + 2e + 1/2 O_2 nbsp;→ H_2O (l) ; nbsp; nbsp; nbsp; nbsp;E^o = + 1.23 V Fe^2+ + 2e → Fe(s) ; ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Gibb's Energy and Nernst Equation

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

-161 kJ/mol

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-1522 kJ/mol

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-76 kJ/mol

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

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}

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}

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^{⊙}

Join BYJU'S Learning Program