Answer to Question #341753 in General Chemistry for Beisnkageu

Question

Balance the REDOX reaction

1. K₂Cr₂O7+ H2O ------- SO₂+ KOH +Cr₂03

2. FeSO4 +KI + KCLO3 ------- Fe(OH)2 )K2 SO4 + 12

Answer 1

1) $2K_2Cr_2O_7 +2 H_2O + 3S = 3SO_2 + 4KOH + 2Cr_2O_3$

$3S^0 - 12e^- = 3S^{4+} - oxidation,$  S is reducing agent

$4Cr^{4+} +12 e^-=4Cr^{3+} - reduction$ , $K_2Cr_2O_7$ is oxidizing agent

2) $3 FeSO_4 + 2 KClO_3 + 12 KOH → 3 K_2FeO_4 + 2 KCl + 3 K_2SO_4 + 6 H_2O$

$2Cl^{5+} +12 e^- = 2Cl^-, reduction$ , $KClO_3$ is oxidizing agent

$3Fe^{2+} -12e^-=3Fe^{6+}, oxidation$ , $FeSO_4$ is reducing agent