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 0.500 M, and the potential on its electrode is measured to be 0.065 V higher than the other electrode. What is the concentration of the other solution? Enter your answer in M.

Answer 1

Answer:

0.003 M

Explanation:

The difference potential on an electrochemical cell can be calculated by the Nernst equation, which is, in its modified form:

ΔE = (-0.0592/n)*logQ

Where n is the number of electrons involved in the reaction, and Q is the coefficient of the reaction. The reaction on that is

Zn⁺²(cathode) → Zn⁺²(anode)

And 2 electrons are being replaced on this equation to the solution of ZnSO₄.

Q = [Zn⁺²(anode)]/[Zn⁺²(cathode)]

To be spontaneous, the cathode must have a higher value of reduction potential, because the reduction must happen at it, and the oxidation must happen in the anode (electrons flow from anode to cathode). Thus, [Zn⁺²(cathode)] = 0.500 M.

0.065 = (-0.059/2)*log([Zn⁺²(anode)]/0.500)

log([Zn⁺²(anode)]/0.500) = -2.2034

[Zn⁺²(anode)]/0.500 = $10^{-2.2034}$

[Zn⁺²(anode)]/0.500 = 6.26x10⁻³

[Zn⁺²(anode)] = 0.003 M