Question

Find the Half-Reactions of the following equation in a BASIC solution, Balance the basic equation, and indicate which substances are oxidizing and reducing agents:

CrI3(s) + H2O2(aq) = CrO4^2-(aq) + IO4-(aq)

Answer 1

Answen (+3) cro I (-1) (+6) 2 (S) + H2O (aq) - cro 4 lag) + Io (ar) to 120 2 Here 1202 reduced Half reactions: oxidation half cm 3t+80H -Be croir t 4H2O ... (1) Reduction half (-2) 16+30M H2O2 tot 43 H2O +20 H2O2 + 20 - 209 20 4 2 H2O +30H -tid To balence the equation, the eakation Gris multiplied by a and quation -in is multiplied by 3 and 2 cr3t+160t - be z eroa +8 H₂O 3*1202 +be - bote 2 cr3+ + 3 H2O2 + 160H 2 Crop +8H₂0 +6012 = 2003 + + 32 0 2 +1004 - 20004² +8ti2o ----- According to the above calculation, the balance equation is- zer Iz + 3 H2oz + LOGH - 2 croq + 610 4 +8H₂O In the equation-liv), the number of hychogen and oxygen atoms are still not balanced. so, the final overall balence equation iS,- 2 cro I 3 + 27 H202 + locH = 2 CrO 4 +61O4 +32 Hoo

2 In equation-i cr3+ (oxidation state,+3) is oxidised to croq²- (oxidation state, +6). Hence crit is an reduceing agent. And In equation -lii H2O2 (oxidation state, -1) reduced to H2O loxidation state, -2). Hence H2O2 is an oxidising agent.