Question

A concentration cell is built using two La/La3+ half-cell electrodes. Given that the cell potential, Ecell for this concentration cell is 0.136 V and the more dilute half-cell has a La3+ concentration equal to 7.8 x 10-8 M, what is the molar concentration of the more concentrated La3+ solution? a) [La3+]conc = 1.0 x 10-14 M b) [La3+]conc = 1.5 x 10-5 M c) [La3+]conc = 7.7 x 10-5 M d) [La3+]conc = 0.61 M e) [La3+]conc = 1.6 M

please show detailed work! i am trying to learn how to do this for a test

Answer 1

Answer: a) [La3+]conc = 1.0 x 10^-14 M

For concentration cell E⁰_cell = 0 because both electrodes are same.

so E_cell = -0.0592/n x logQ is the formula

where n = 3 (La3+); Q = anode / cathode concentrations; given E_cell = 0.136 V

anode = more concentrated electrode; cathode = less concentrated electrode

substitute these values in E_cell = -0.0592/n x logQ

0.136 V = -0.0592 / 3 x log[anode / 7.8 x 10-8]

use maths,

log[anode] = -14

[anode] = 10^-14

molar concentration of the more concentrated La3+ solution = 1.0 x 10^-14 M

Hope this helped you!

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