JEE Advanced2013ChemistryElectrochemistryActual
The standard reduction potential data at 25 ° C is given below. E ⁡ ο Fe 3 + Fe 2 + = + 0 · 7 7 V ⁡ ; E ⁡ ο Fe ⁡ 2 + Fe ⁡ = - 0 · 4 4 V ⁡ E ⁡ ο Cu ⁡ 2 + Cu = + 0 · 3 4 V ⁡ ; E ⁡ ο Cu + Cu = + 0 · 5 2 V E ⁡ ο O 2 g ⁡ + 4 H ⁡ + + 4 e ⁡ - → 2 H ⁡ 2 O = + 1 · 2 3 V ⁡
Options
- Aa-r;b-q;c-s;d-p;
- Ba-q;b-p;c-r;d-s;
- Ca-s;b-p;c-r;d-q;
- Da-r;b-s;c-p;d-q;
Correct answer
D. a-r;b-s;c-p;d-q;
Step-by-step solution
P.) Fe 3 + + e - ⟶ Fe 2 + ; Δ G I o = - 1 F 0 . 7 7 Fe 2 + + 2 e - ⟶ Fe ; Δ G II o = - 2 F - 0 . 4 4 ---------------------------------- Add Fe 3 + + 3 e - ⟶ Fe ; Δ G III o = - 3 F E o Fe 3 + / Fe - 3 F E o Fe 3 + / Fe = - 0 . 7 7 F + 0 . 8 8  F = 0 . 11 F E Fe 3 + / Fe o = - 0 . 1 1 3 = - 0 . 0 3 7 ≅ - 0 . 0 4  V Q.) 2 H 2 O ⟶ O 2 g + 4 H (aq) + + 4 e - (Oxidation half reaction) E ° = - 1 . 23   V O 2 g + 2 H 2 O + 4 e - ⟶ 4 OH (aq) - (redu