## Answers

dihtim — + C10; +7 + Fe 0 reduxtin C1 -1 Fe3+ +3 Step: 1 Oxidation half reaction: Fe — Reduction half reaction: C10 Fe+ 01 - Step: 2 Balancing oxidation half reaction: Fe - Fe** Balancing charge Fe —— Fe* +3e Step: 3 balancing reduction half reaction: C10 — a Balancing C1 C10 0 1 Balancing 0: 00 —C1" + 4H,0 Balancing H: CIO: + 8H* —— C1° +4,0 Balancing charge: C102 + 8H+ + 80 – 01 +4H, O Step 4: Equalizing the no of electron lostin oxi dati on half reaction and no of electrons gainedin reduction half reaction. (C107 + 8H* +8 01 +41,0 )x3 (Fe —— Fe* +3e )x8 Step 5: Adding both the half reactions 8Fe 3C10; + 24H+ + 24 — 3C10; + 24H+ + Fe — 8F2+ 24e- 3C1° +12H2O 8Fe +3C1° +12H,O

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