Dr. Sangeeta Khanna Ph.D 1 CHEMISTRY COACHING CIRCLE Macintosh:Users:sahil:Desktop:Sub-Electrochemistry-Answers.doc Electrochemistry (Subjective Assignment) Answers 1. (a) 0 oxidation E of Cu + , Fe 2+ , Hg + , Br – are given. 0 oxidation E of Fe 2+ Fe 3+ is highest. Hence Fe 2+ is more oxidized & Fe 2+ is strong reducing agent. Since Fe 2+ is better reducing agent than Cu + , therefore Fe 2+ will reduce Cu 2+ → Cu + (b) I – is strong reducing agent. It reduces Cu 2+ into Cu + . Therefore reaction of Cu 2+ & I – produces Cu 2 I 2 rather than [CuI 4 ] 2– . Cl – does not reduce Cu 2+ into Cu + . Hence, reaction of Cu 2+ & Cl – leads to formation of [CuCl 4 ] 2– 2. (i) Cl 2 & H 2 are given off at anode & cathode respectively 2Cl – –2e – → Cl 2 2H + + 2e – → H 2 (ii) H 2 & O 2 are given off when Cl – conc. is very low, water molecules begin to discharge at anode, while solution remain acidic, H + are likely to be discharged in preference to H 2 O molecules but eventually letter molecules will be discharged. At anode H 2 O (l) → 2H + + 2 1 O 2 (g) + 2e – At cathode 2H 2 O + 2e – → H 2 + 2OH – (iii) H 2 & O 2 are given off at anode & cathode. 3. (a) Fe 3+ + Ce 3+ → Ce 4+ + Fe 2+ V 60 . 1 76 . 0 E 0 cell + = = -0.8 4 V So Fe 3+ will not oxidize Ce 3+ (b) V 0.82 - V 1.36 - V 54 . 0 E 0 cell = = I 2 does not displace Cl 2 from KCl (c) 2FeCl 3 + SnCl 2 → SnCl 4 + 2FeCl 2 r 0.61, 15 . 0 76 . 0 E 0 cell = + = so this reaction occur 4. (a) H 2 →2H + + 2e – 2 2 0 H 2 H pH ] H [ log 2 059 E E + + − = =0.0–0.0591 log [H + ] 0.25 = 0.059 pH pH = 0.25/0.0591 = 4.23 (b) ] H [ log 059 . 0 E E 0 H 2 H + + − = 0.3 = 0.0591 pH pH = 5.076 5. When MnO 2 will be used up in cathodic process, the dry cell will stop to produce current. Cathodic process 2MnO 2 +Zn 2+ + 2e – → ZnMn 2 O 4 Eq. mass of MnO2 = 87 1 87 state oxidation in change mass molecular = = From 1 st law of electrolysis 96500 t i E × × = ω 87 10 4 8 96500 t 3 × × × = −