Question

An electrochemical cell is constructed such that on one side a pure lead electrode is in contact with a solution containing Pb2+ ions at a concentration of 0.002 M. The other cell half consists of a pure nickel electrode that is immersed in a solution of Ni2+ ions having a concentration of 0.4 M. Given that the standard electrode potentials for lead and nickel are -0.126 and -0.250 V, respectively, at what temperature will the potential between the two electrodes be -0.012 V?

Answer 1

-490.7 K

Step-by-step explanation:

Given:

[Ni^2+]= 0.4 M

[Pb^2+]=0.002 M

∆V= -0.012 V

VNi= -0.250V

VPb= -0.126V

F= 96500 C

R= 8.314 JK-1 mol-1

n= 2

From

T= -nF/R [∆V-(VNi-VPb)/ln [Pb2+]/[Ni2+]]

T= 2(96500)/8.314[ (-0.012) -(-0.250) - (-0.126))/ln[0.002]/[0.4]

T= 23213.856(0.112/(-5.298))

T= -490.7 K