Question

For the following unbalanced reaction, A and X represent generic elements, H and Orepresent hydrogen and oxygen, respectively. Please answer the following questions that relate to this reaction. A9H2002(aq) + X2072-(aq) → A9H1604(aq) + X3+(aq) 1. Assign oxidation numbers to all elements in the above reaction (A, X, H, O), and please show your calculations 2. Label the substance that is the oxidizing agent and the substance that is the reducing agent 3. Separate the oxidation and reduction half-reactions, and label which step is oxidation and which is reduction 4. Balance the redox reaction that takes place in acidic solution using the half- reaction method, and clearly show all steps of the process to receive full credit

Answer 1

1) Oxidation number of H = +1 (common)

Oxidation number of O = -2 (common)

Let the oxidation number of the unknown element be x.

AqH₂o O2 : 92 + 20(+1) + 24-2) = 0 + = x= -16 a - logo X2072 : .. 236 +7(-2) = -2 a x = +6 Aq H16 On : 9x + 16(+1) + "(-2) -0 = 2 = 8 colo X3+ :+3 (change oxidation number) increase in oon. = Oxidation I +6 -16/2 Aq H₂o O₂ + x₂0₂ 2- -8/9 Aq Hibou + X3+ Reducing agent decrease in oon = reduction 2- oxidising agent XzO² de Oxidation Half-equation : Aq H₂o O₂ Aq His Ou +86- To balance o, add H₂O

Aq H₂o O₂ + 2 H₂O da Hib Ou +86- To balance H. add Ht Aq H₂O + 2 H₂O - Aq Hic Oy + 8H7 + 8e- Reduction Half equation : 2- X₂07 + bé → 2 x 34 To balance o, add H₂O : X 20 2- + bé → 2X3+ + 7 H20 To balance H, add Ht. 3+ 2- X₂07 + 14 H+ +be → 2X + 7 H₂O Multiply oxidation half-equation by 3, and reduction half-equation by 4; add them together 3 Aq H2o O₂ + 6H20 + 4X20 2 - - - 2 ht + 246 3 Aq H16 Ou + 34-44 +246 +8x3+ + 28 H20 22 3 AqHz Q (aq) + 47203(aq) + 32 H+ (aq) → 3+ 3 AgH6Oy (aq) + 8 x** (aq) + 22 H20 (1)