plz help answer what u can i will ask the other questions as a follow up quesiton is the same thread \ ty QUESTION 14Continue with the last three questions, calculate the equilibrium constants K of the redox reactions at 25°C.4.6 x 10161.040.041.4 x 1016QUESTION 15Continue with the last question, based on your calculation, which direction does the reaction proceed spontaneously under thereaction condition?O Toward productcannot be determined.O The reaction is in equlibriumO Toward reactant QUESTION 11The following 5 questions refer to the redox reaction shown below at 298K :Sn (s) + CuSO4 (s) → SnSO4 (aq) + Cu (s)write the electrode half reactions.Anode: Sn (s) +2e Sn2+ (aq); cathode: Cu (s) +2eAnode: Cu (s) +2e → Cu2+ (aq); cathode: Sn (s) +2eQUESTION 12-Anode: Cu²+ (aq) +2e → Cu (s); cathode: Sn²+ (aq) +2eAnode: Sn (s) - 2eSn2+ (aq); cathode: Cu²+ (aq) +2eOPt | Sn(s), Sn²+ (aq) || Cu2+ (aq), Cu(s) | PtO Sn²+ (aq) | Sn(s) ||Cu(s) | Cu²+ (aq)Sn(s) | Sn2+ (aq) || Cu²+ (aq) | Cu(s)QUESTION 13→Continue with the last question, the corresponding cell notation isO Sn(s) | Cu2+ (aq) || Sn²+ (aq) | Cu(s)Cu2+ (aq)Sn²+ (aq)O 0.197-0.197→ Sn (s)Cu (s)Continue with the last two questions, calculate stand cell potential Eºcell.0.477-0.477

According to guidelines we are supposed to answer only first three subparts if you want rest of the parts to be solved then please post them as a separate questions. In an electrochemical cell, chemical energy is covered into electrical energy. Oxidation takes place at anode and anode is negative. Reduction take place at cathode and cathode is positive.Given reaction, Sn(s) + CuSO4(s) --> SnSO4(aq) + Cu(s) In this reaction Sn(s) looses two electron and becomes Sn2+(aq) this means it is oxidation and oxidation takes place at anode and Cu2+(aq) gain two electrons and becomes Cu(s) this means reduction and reduction takes place at cathode. Anode: Sn(s) - 2e- --> Sn2+(aq) Cathode: Cu2+(aq) + 2e- --> Cu(s) So, correct option is fourth option Cell notation In cell notation, oxidation is written at left the species in different states are speared by | In the middle we have salt bridge which is represented by || Oxidation is written on right and species in different states are represented by | Anode: Sn(s)|Sn2+(aq) Cathode: Cu2+(aq)|Cu(s) Cell notation Sn(s)|Sn2+(aq)||Cu2+(aq)|Cu(s) So, correct option is fourth option E°= E°reduction-E°oxidation Here, all the potential are reduction potential E°= E°(Cu2+(aq)|Cu(s))-E°(Sn2+(aq)|Sn(s)) Standard values E°(Cu2+(aq)|Cu(s))=0.337V E°(Sn2+(aq)|Sn(s))=-0.14V Substituting values, E°=0.337-(-0.14) E°=0.447 So, correct option is first one