Question

Electrochemical cells can be used to measure ionic concentrations. A cell is set up with a pair of Zn electrodes, each immersed in ZnSO4 solution. One solution is created to be exactly 0.400 M, and the potential on its electrode is measured to be 0.045 V higher than the other electrode. What is the concentration of the other solution? Enter your answer in M.​

Answer 1

for the first electrode

E1= Eo + (RT/zF) ln [Zn+2]

given

[Zn+2] = 0.4

so

E1 = Eo + (RT/zF) ln 0.4

now

for the second electrode

E2 = Eo + (RT/zF) ln [Zn+2]

now

given

E1 - E2 = 0.045

now

E1 - E2 = Eo + (RT/zF) ln 0.4 - Eo - (RT/zF) ln [Zn+2]

E1 - E2 = (RT/zF) ln 0.4 / [Zn+2]

so

0.045 =( RT/zF) ln 0.4 / [Zn+2]

now

Zn+2 + 2e- --> Zn

so

z = 2 as two electrons are transferred

so

0.045 = (8.314 x 298 / 2 x 96485 ) ln 0.4 / [Zn+2]

[Zn+2] = 0.012

so

the concentration of the other solution is 0.012 M