We are being asked to determine which change will result to increase the voltage of the cell:

**Al(s) | Al ^{3+}(aq, 1.00 M) || Cu^{2+}(aq,1.00 M) | Cu(s)**

In the Nernst Equation, we will **calculate for Q** where Q is equal to **[products]/[reactants]**.

**The Nernst Equation at 25****°C:**

**Rewrite the Nernst Equation:**

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

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

E°_{cell} = standard cell potential

n = number of e^{-} transferred

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

We have to write the overall reaction and determine the products and reactants:

Which of the following changes will increase the voltage of the cell Al(s)|Al^{3+}(aq, 1.00 M)||Cu^{2+}(aq,1.00 M)|Cu(s)?

I. Increasing the surface area of the Al electrode

II. Decreasing the concentration of Al^{3+}(aq) to 0.001 M

III. Decreasing the concentration of Cu^{2+}(aq) to 0.001 M

a. I and II`

b. I and III

c. I only

d. II only

e. III only

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