We are asked to calculate for the **E _{cell} for the reaction**.

**Mg(s) I Mg ^{2+}(aq)(0.15 M) II Au^{+}(aq)(0.3 M) I Au(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.

What is the E_{cell} for Mg(s) I Mg^{2+}(aq)(0.15 M) II Au^{+}(aq)(0.3 M) I Au(s) ? Assume that the number of electrons transferred in the reaction is n = 2.

Mg^{2+} + 2e^{-} → Mg E°_{red} = -2.36 V

Au^{+} + e^{-} → Au E°_{red} = +1.69 V

1. 4.04 V

2. 4.08 V

3. 3.2 x 10^{-2} V

4. 4.03 V

5. 4.01 V

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