We are asked to calculate the **cell potential o****r emf** of the given voltaic cell at the given concentrations, with the overall reaction:

**3Ce**^{4 + }** (aq) + Cr(s) ****→**** **** 3Ce**^{3 + }** (aq) + Cr**^{3 + }** (aq)**

We will use the **Nernst Equation** to calculate the cell potential with the given conditions. The Nernst Equation relates the concentrations of compounds and cell potential.

$\overline{){{\mathbf{E}}}_{{\mathbf{cell}}}{\mathbf{=}}{\mathbf{E}}{{\mathbf{\xb0}}}_{{\mathbf{cell}}}{\mathbf{-}}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{0592}}{\mathbf{n}}\mathbf{\right)}{\mathbf{logQ}}}$

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 determine the **E°**** _{cell }**first and as well as the

To do so, we need to do the following steps:

**Step 1**. Identify the *reduction half-reaction (cathode) and the oxidation half-reaction (anode) ***Step 2**. Balance the *number of electrons transferred **by adding the reduction half-reaction and oxidation half-reaction to get the given overall reaction Step 3. Determine the half-cell potentials (refer to the Standard Reduction Potential Table at 25°C) to Calculate *

A voltaic cell utilizes the following reaction and operates at 298 K:

3Ce^{4 + } (aq) + Cr(s) → 3Ce^{3 + } (aq) + Cr^{3 + } (aq).

What is the emf of this cell when [Ce^{4 + } ] = 2.6 M , [Ce^{3 + } ] = 0.16 M , and [Cr^{3 + } ] = 1.6×10^{−2} M ?

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