We are asked to calculate for the cell potential (E_{cell}) for the voltaic cell below:

**Zn(s), [Zn ^{2+}] = 4.50M] || Cu(s), [Cu^{2+}] = 0.0120M**

We can rewrite the **cell notation** as:

**Zn(s) | Zn ^{2+}, (4.50 M) || Cu^{2+}, (0.0120M) | Cu(s)**

Recall that the** Nernst Equation** relates the concentrations of compounds and cell potential.

$\overline{){{\mathbf{E}}}_{{\mathbf{cell}}}{\mathbf{=}}{\mathbf{E}}{{\mathbf{\xb0}}}_{{\mathbf{cell}}}{\mathbf{-}}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{05916}\mathbf{}\mathbf{V}}{\mathbf{n}}\mathbf{\right)}{\mathbf{}}{\mathbf{log}}{\mathbf{}}{\mathbf{Q}}}$

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

E°_{cell} = standard cell potential

n = mole e^{-} transferred

Q = reaction quotient = products/reactants

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

*Step 1:** **Identify the anode and the cathode in the reaction and write the overall reaction*** Step 2: **Calculate the cell potential of the reaction.

Given the voltaic cell below, find the cell potential.

Zn(s), [Zn^{2+}]=4.50M] || Cu(s), [Cu^{2+}]=0.0120M

Cu^{2+}(aq) + 2e^{–} ⇌ Cu(s) E° = 0.34 V

Zn^{2+}(aq) + 2e^{–} ⇌ Zn(s) E° = -0.76 V

A) 1.23 V

B) 1.10 V

C) 1.18 V

D) 1.02 V

E) 1.08 V

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