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.
Ecell = 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 (Ecell) using Nernst Equation
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 ?