Question

Enter your answer in the provided box. What is the emf of a cell consisting of a Pb2+ / Pb half-cell and a Pt/" /H, half-cell if [Pb2+] = 0.57 M, [H] =0.012 M and PH, = 1.0 atm ? 0.024 V

Please help me!! Answer is not 0.024..

Answer 1

Standard reduction potential of Pb^{​​​​​2+}/Pb is -0.13 V

Standard reduction potential of Hydrogen electrode is 0 V

According to Electrochemical series, lead electrode acts as anode and hydrogen electrode is the cathode.

Variation of electrode potential with concentration of reactants and products is given by Nernst equation.

Cell n represente an 142 Ehcto chamic eachon h witte an Ph (007) 5) Pb+2 (0.012 M) ph P2e No 4 electmm transtered 2 E E 0-C0-13) 012v Calude Call Ecall Coll O-STM X1 a Adutr O13V Ecall 2 MNTVD02y V 0-13 0106 1 Ph H pb 2117 PL O.o12 -0.13-tog a012) 0 los Eall T -O.13-O O59 N36V 0.024 V

If Lead is taken as anode and Hydrogen electrode is as cathode, emf will come only +0.024 V.

If we take reversely, ie Lead as cathode and Hydrogen electrode as anode, emf will come -0.024 V