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.

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 =

[Zn⁺²(anode)]/0.500 = 6.26x10⁻³

[Zn⁺²(anode)] = 0.003 M

the increase of protons   hope this

the answer to this question is 3.