# 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 ?

###### FREE Expert Solution

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

3Ce4 +  (aq) + Cr(s)   3Ce3 +  (aq) + Cr3 +  (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.

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

Ecell = cell potential under non-standard conditions
cell = standard cell potential
n = number of e
- transferred
Q= reaction quotient = [products]/[reactants]

We have to determine the 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
cell.
Step 4. Calculate
cell potential (Ecellusing Nernst Equation ###### Problem Details

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 ?