Question

Piece #1 (1/2 pt) Consider the following reaction (unbalanced): MnO2 (s) + H3AsO3(aq) -Mn²+ (aq) + H2AsO(aq) Split the reaction into two half-reactions. Oxidation half reaction: H:AsO3(aq) → H3 AsO, (aq) Reduction half reaction: MnO2 (s) Mn (aq) Piece #2 (1/2 pt) Consider the following half-reaction: MnOz(s) - Mn2+ (aq) Balance everything but oxygen and hydrogen atoms. MnO2 (s) + 2e Mn (aq)

Answer 1

H 3-2 +5-2 s Step 1 : write the unbalanced equation 8 mnogest Hz As 03 (og mun & GAS 04 (99) step & or Assign oxidation numbers : to ar, th 32 Mug & H As Oy - & the As a Step 3: Identify the oxidation and reduction helfs and write them separately. oxidation : 1h As Oz - Hl. As a Reduction & th e Step 4 - Balance the sleepsons : - 08 . As – M As 04 & 2é Ro tintzen dira Step 5: Balance the charge by adding stt gous! or de ASO . HASO4 + 20 & 24+ Ro Muoz & 26 + 4H+ Muht Step 6 e Balance the oxygen alony by adding thor 08- H₂. As Og + tho H₂ As Oq & 26 + 2H+ +1 3-2 H +5.2 Ro muo, & ze + 4H+ Mut + 2460

Step 7 & Make electron goin equivalent to electron lost, as the e-gain and cost in our reduction and oxidation hall is equal than adding the equations, we get :- H₂ As O₂ t. Mno, & 2H + cag (S) cage H. As 04 & cab) . Muht & Cage