When cadmium metal is reduced Cu^2+ in solution Cd^2+ forms in addition to copper metal. If ∆G° =143kJ, Calculate K at 25°C. What is the E^0 cell of a voltaic cell that uses this reaction
To calculate K at 25°C, we can use the relationship between ∆G° and K:
∆G° = -RT ln K
where R is the gas constant (8.314 J/mol·K) and T is the temperature in Kelvin (25°C = 298 K).
We are given ∆G° = 143 kJ. To be consistent with the units of R, convert ∆G° to J:
∆G° = 143 kJ * 1000 J/kJ = 143,000 J
Now we can calculate K:
143000 J = - (8.314 J/mol·K) (298 K) ln K
143000 J = - 2471.972 J/mol·K ln K
Now, divide both sides by -2471.972:
-ln K = -58.455
Now, take the exponential of both sides to solve for K:
K = e^(58.455) = 4.37 x 10^25
Now, we can find E^0 cell using the relationship between ∆G° and E^0 cell:
∆G° = -nFE^0_cell
where n is the number of moles of electrons transferred (in this case it's 2, as Cd is reduced from Cd^2+ to Cd and Cu^2+ is reduced to Cu) and F is Faraday's constant (96,485 C/mol).
∆G° = 143000 J = -(2)(96485 C/mol)E^0_cell
Now, solve for E^0_cell:
E^0_Cell = 143000 J / (2 * 96485 C/mol) = 0.74 V
The E^0 cell of the voltaic cell that uses this reaction is 0.74 V.