We are asked to calculate the value of E for a cell in which [Cu^{2+}] = 1.0 x 10^{-3} M and [Zn^{2+}] = 1.0 M.

Zn(s) + Cu^{2}^{+}(aq) ⇌ Zn^{2+}(aq) + Cu(s) E°_{cell }= 1.10 V

We will use the **Nernst Equation** to calculate the cell potential. The Nernst Equation relates the concentrations of compounds and cell potential.

$\overline{){{\mathbf{E}}}_{{\mathbf{cell}}}{\mathbf{=}}{\mathbf{E}}{{\mathbf{\xb0}}}_{{\mathbf{cell}}}{\mathbf{-}}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{0592}\mathbf{}\mathbf{V}}{\mathbf{n}}\mathbf{\right)}{\mathbf{}}{\mathbf{log}}{\mathbf{}}{\mathbf{Q}}}$

E_{cell} = cell potential under non-standard conditions

E°_{cell} = standard cell potential

n = number of e^{-} transferred

Q= reaction quotient = [products]/[reactants]

We're going to calculate the E_{cell} using the following steps:

The value of E°_{cell} for the reaction Zn(s) + Cu^{2}^{+}(aq) <=> Zn^{2+}(aq) + Cu(s) is 1.10 V. What is the value of E for a cell in which [Cu^{2+}] = 1.0 x 10^{-3} M and [Zn^{2+}] = 1.0 M?

a. 1.01 V

b. 1.19 V

c. 0 V

d. -1.01 V

e. -1.19 V

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