Question 25.5: The value of E°cell at 25.0°C for the equation Zn(s) + Cu^2+......
The value of E^{\circ}_{cell} at 25.0°C for the equation
Zn(s) + Cu^{2+}(aq) ⇋ Cu(s) + Zn^{2+}(aq) (25.11)
is 1.10 V. Calculate the equilibrium constant at 25.0°C for this equation.
From Equation 25.10, we have
ln\ K =\frac{ν_{e}E^{\circ}_{cell}}{0.02570\ V} \qquad (at\ 25.0^{\circ}C) (25.10)
ln\ K =\frac{ν_{e}E^{\circ}_{cell}}{0.02570\ V}For each mole of Zn(s) in the oxidation-reduction reaction described by Equation 25.11, two moles of electrons are transferred to a mole of Cu^{2+}(aq). Thus, the value of ν_{e} is 2. Therefore,
ln\ K =\frac{(2)(1.10\ V)}{0.02570\ V}=85.6and
K =\frac{[Zn^{2+}]eq}{[Cu^{2+}]eq} =e^{ 85.6} = 1.50 × 10^{37} ≈ 2 × 10^{37}The very large value of K means that at equilibrium, the ratio of [Zn^{2+}] to [Cu^{2+}] is very large or, in other words, that the value of [Cu^{2+}] at equilibrium is very small. (The subscript eq on a concentration term denotes a chemical equilibrium value of that concentration.)