Question

What is the emf of the cell Pt, H2(Ibar) HCl(ml)//HC/(m2)/Pt, H2(lbar) If ml = 0.2 and m2 = 0.3 Select one: O a. 0.03 V b. -0.01 V O c. 0.01 V d. -0.10 V e. 0.10 V Next

Answer 1

DAT Answer:- 0.0IV The given cell representation is, Cleft) (Right) Pt, H2 C1 bar)Hacmi) HCI CM2) I PE ,H2Cl bur) According to the convention used for cell representation, the electrode present at left is the anode at which Oxldation takes place and the electrode present at right is the cathode act which reduction takes place. : Half cell reaction at anode is, % H₂ Ht (mi) tle Half cell reaction at cathode is, Ht (mastlé V₂ H2 :. Net cell reaction is the combination of reaction at anode and cathode Canade): the sht (mité (cathode): Htcm.) the Net cell: Hima) > Hom) reaction + . Reaction Quatient [mi] 1m2 [mi] [m2]

DATE The emf of the cell can be calculated by using the Nernst's equation as given below: Ea F° – 2.303 RT log Q nF where, E emf of cell standard emf of cell R Universal gas constant ( 8.3144 J kimon) Ia Temperature (Room Temp, 25°C = 298K) f→ Faraday constant (964850) in number of electrons involved in the cell reaction Q Net cell reaction quotient. E 2°- 2:303 X8-31217 298 log Q nx 96585 во G 0.059 1 logo n for the given reaction, anode and cathode both are Hydrogen electrodo 1 And, Standard electrod potential for Hydraden electrode Pos = 0.00 V és S Fandard cell emf CEO) = 0.00 V from net cell reaction, we get an idea, that in the cell reaction only le involvement

And, from ear , Q = [mi] Cm] We have, m=0.2 and m2 70.3. 0.2 0.3 Now, putting the values of E, n and Q in egr 0, we get E= 0 0.059 1 log/0.2 0.3 -0.0591 (-0-1760) E = +0.0104 V TE- 0.01 V : Emf of given cell reaction is o.olv