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 = [tex]10^{-2.2034}[/tex]
[Zn⁺²(anode)]/0.500 = 6.26x10⁻³
[Zn⁺²(anode)] = 0.003 M
