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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.

Respuesta :

IthaloAbreu

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:

Ī”E = (-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

ZnāŗĀ²(cathode) ā†’ ZnāŗĀ²(anode)

And 2 electrons are being replaced on this equation to the solution of ZnSOā‚„.

Q = [ZnāŗĀ²(anode)]/[ZnāŗĀ²(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, [ZnāŗĀ²(cathode)] = 0.500 M.

0.065 = (-0.059/2)*log([ZnāŗĀ²(anode)]/0.500)

log([ZnāŗĀ²(anode)]/0.500) = -2.2034

[ZnāŗĀ²(anode)]/0.500 = [tex]10^{-2.2034}[/tex]

[ZnāŗĀ²(anode)]/0.500 = 6.26x10ā»Ā³

[ZnāŗĀ²(anode)] = 0.003 M

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