The cell voltage can be calculated using the Nernst equation:
Ecell = E°cell - (RT/nF)ln(Q)
where:
- E°cell is the standard cell potential
- R is the gas constant (8.314 J/K·mol)
- T is the temperature in Kelvin (25.0°C = 298.15 K)
- n is the number of electrons transferred in the balanced redox reaction (in this case, n = 6)
- F is the Faraday constant (96,485 C/mol)
- Q is the reaction quotient, which can be calculated as [Cr²⁺]^2[Ca²⁺]^3/[Cr³⁺]^2
First, let's calculate the standard cell potential, E°cell. We can do this using the standard reduction potentials for the half-reactions involved:
Cr³⁺(aq) + e⁻ → Cr²⁺(aq) E°red = -0.407 V
Ca²⁺(aq) + 2e⁻ → Ca(s) E°red = -2.870 V
The overall reaction is the sum of the two half-reactions, so we can add their standard reduction potentials to get the standard cell potential:
E°cell = E°red,cathode - E°red,anode
E°cell = 0.407 - (-2.870) = 3.277 V
Now we can plug in the values into the Nernst equation and solve for Ecell:
Ecell = 3.277 V - (8.314 J/K·mol)(298.15 K)/(6 mol)(96,485 C/mol)ln((1.96 M)^3/(4.01 M)^2)
Ecell = 3.277 V - 0.059 V
Ecell = 3.218 V
Therefore, the cell voltage under these conditions is 3.218 V (rounded to 3 significant digits).