Question

please answer number 8 and 12 and explain. please answer number 12 in units of kJ/mol

Question 8 5 pts Consider the following table of standard electrode potentials for a series of hypothetical reactions in aqueous solution: Reduction Half- E® (v) Reaction A*(aq) + e- A(s) 1.57 B2+(aq) + 2 e- B(s) 1.11 c+(aq) + e- → C(s) 0.83 D3+ (aq) + 3e D(s) -0.47 E* (aq) + e → E(s) -1.95 What is the standard cell potential for the following cell reaction? A+ (aq) + E(s) – A(s) + E+ (aq) 2.78 v -0.38 V 3.52 v

Answer 1

Ans & The given cell reaction is A+ (aq) + E(3) A(s) + Et (ag) from the cell reaction, the oxidation number of A change from th too and E change from 0 to +1 loss of electrow is called oxidation and oxidation occurs at anode. So, we can write the cell reaction as Oxidation half cell :- E(s) + (aq) + é (Anode)

Reduction half cell: Atlas) té A (s). (Cathode) Standard cell potential, Erell is given by E cell = Ecathode - Eanode = 1.57-1-1.95 V = 1.57+1.950 = 3.52V Jo, the correct option is ② 3.52v. 12 Ans - The gurn cell reaction o r y2+ (aq) + 2 2(5) Y(s) +22+ (aq) Oxidation number of a change from 0 to + I and that of y change from ta to O. Jo, a is oxidised and Yis reduced at cathode . Half all reactions are : Oxidation half: 22(5) 22+ (aq) + 2e Reduction half - yat (aq) + aé - Y(s). Ecell = E cathode -Eanode - 1.12 -0.87 V = 0.25 V Again, on is given by G=-nf Ecell where f = 96487 cmolt and n=2 AG =-2x0.25 Ux96487 cm = - 482 43.5 J mot!

N, 0G = 48.243 kJ/md. So standard free energy change, ag is 48.243 kJ/mol