Problem: A voltaic cell utilizes the following reaction and operates at 298 K:3Ce4 +  (aq) + Cr(s)  →  3Ce3 +  (aq) + Cr3 +  (aq).What is the emf of this cell when [Ce4 +  ] = 2.6 M , [Ce3 +  ] = 0.16 M , and [Cr3 +  ] = 1.6×10−2 M ?

We are asked to calculate the cell potential or 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.

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

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 anode (oxidation)and cathode (reduction) reactions and the number of electrons transferred (n). 

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 E°_cell.
Step 4. Calculate cell potential (E_cell) using Nernst Equation