Question

please explain with details

sen peroxide is a colorless liquid at room temperature. It is soluble in water because of its unique structure. Figure 1 shows the structure, bond angles, and bond lengths in the molecule. Hydrogen pigure 1 shows Figure 1: Structure of hydrogen peroxide OLS 147.4 pm 1 0 9 95.0 pm .0pm OH 94.8 When heated hydrogen peroxide forms H20 and 02. The reaction is rapid in the presence of a catalyst like potassium iodide. Figure 2: Decomposition reaction of H2O2 and standard free energies of formation of reagents KT 2H2O2(g) → H20(g) + O2(g) AG° -120 kJ/mol -230 kJ/mol 0 KMnO4 reacts with H2O2 in the presence of an acid to also produce O2. Figure 3: Half-Cell Potentials E°(V) Half-Reaction 0.0 +1.51 +1.77 2 H1+ + 20 ==> H₂ Mn0,1- + 8 1+ + 5 ==> Mn2+ + 4 1,0 0 + 2 H+ + 20 ===2 H 42. Given the data in Figure 2, What is the Free Energy change for the decomposition of H,0, in the presence of KI? A) +10kJ B) -90kJC) -110kJ D) - 350kJ 43. What is the cell potential for the following reaction? $4,02 + 6H+* + 2Mn0, -→ 502 + 2Mn2+ + 8H,0 A) +3.28VB) -0.26 V C) +11.87V D) -5.83V

Answer 1

42)

Firstly, we need to balance the reaction such that the coefficient of H₂O₂ is 1. Note that the original reaction is unbalanced.

Thus, we will have,

H2O2(g) = H2O(g) +1/202(9)

For any reaction, the formula for free energy change of the reaction is,

AG reaction action = AG f products - AGf reactants

Here, The products are 1/2O₂ and H₂O and the reactant is H₂O₂

Thus, putting in the value of free energy of these in the above reaction, we have,

A Greaction = (1/2AG;,, + AG,H,0) - AG%,H,02 =AGreaction = (1/2 x OkJ+(-230k])) - (-120kJ) A Greaction = -110kJ

Thus, the answer will be (C) -110 kJ

43)

We have two half reactions

Mno+ 4H+ + 5 = Mn2+ + 4H20 (1) Eº = +1.51V

02+2H+ + 24 = H202 (2)E° = +1.77V

We first, reverse the second reaction,

H202 = 02 + 2H+ + 2e (3) Eº = -1.77V

Note that the sign of E^o will become negative. Then we multiple the reaction (1) with 2 and the reversed reaction (3) with 5 and add both of them to get the full reaction given

5H202 + 2 MnO2 + 6H+ = 50, + 2Mn2+ + 8H20 E

The E^o will be simply the sum of E^o of the two reactions (1) and reaction (3)

E_o = +1.51 V + (-1.77 V) = -0.26 V