chlorine will oxidizing I as shown Use these data to determine the enthalhy chan
ID: 999164 • Letter: C
Question
chlorine will oxidizing I as shown Use these data to determine the enthalhy change for this reaction. Na_2O_2 rightarrow Na_2O+ 1/2 O_2 +99 KJ -99KJ +198 KJ cannot determine without delta H_f for O_2 Chlorine will oxidize I^- as shown in this equation: Cl_2 + 2l^- rightarrow l_2 + 2Cl^- Use reactivity trends to predict which can be used to prepare Br_2 from Br^-. F + Br^- Cl_2 + Br^- l_2 + Br^- III only I and II only II and III only I, II, and III A buried metal tank will not corrode if it is wired to a metal rod that corrodes more easily, Which combination protects the tank from corrosion? Mg rod and S_n tank Fe rod and Z_n tank Fe rod and Al tank Cu rod and Zn tank Cd^2+(aq) + 2e^- rightarrow Cd E'' = - 0.402 Ni^2+(aq) + 2e^- rightarrow Ni E'' = -0.236 -0.638 v What happen during this reaction? Cd + NiO_2 rightarrow Cd(OH)_2 + Ni(OH) Cadmium metal is oxidized. Cadmium metal is reduced The O atom in H_2O is reduced. The Ni atom is NiO_2 oxidized. The solubility of a salt is water usually increase increasing temperature because the process is exothermic so dissolution occurs faster. exothermic so dissolution occurs faster. exothermic to the higher temperature favors dissolution. exothermic to the higher temperatures favor dissolution. Which reagent, when added to BaCl_2(aq) and filtered, can produce a solution that is essentially HCI(aq)? HC_2H_3O_2 H_2SO_4Explanation / Answer
Solution :-
The reduction potential of the Cl2 and F2 is larger than the Br2 so the F2 and Cl2 can oxidize the Br^- to form the Br2
So the answer is option B = l and ll only
because the higher the reduction potential means good oxidizing agent and lower the reduction potential means the good reducing agent.
Related Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.