Problem: Which of the following changes will increase the voltage of the cell Al(s)|Al3+ (aq, 1.00 M)||Cu2+ (aq, 1.00 M)|Cu(s)?I. Increasing the surface area of the Al electrodeII. Decreasing the concentration of Al3+(aq) to 0.001 MIII. Decreasing the concentration of Cu2+(aq) to 0.001 Ma. I and IIb. II and IIIc. II onlyd. III onlye. I only

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

Al(s) | Al³⁺(aq, 1.00 M) || Cu²⁺(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:

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

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: