We're being asked to calculate the emf generated by the cell when [Al^{3+}] = 4.5×10^{‒3} M and [I^{‒}] = 0.15 M.

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

**Step 1: **Write the two half-reactions and determine the anode and cathode.**Step 2:** Write the overall reaction and determine the number of electrons transferred (n).**Step 3: **Calculate E_{cell} using the Nernst equation.

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

**Step 1: **Write the two half-reactions and determine the anode and cathode.

The standard emf for the cell using the overall cell reaction below is +2.20 V:

The emf generated by the cell when

[Al^{3+}] = 4.5 x 10^{-3} M and [I^{-}] = 0.15 M is _____ V.

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