. A small amount of solid Hg2Cl2 and 25 ml of metallic mercury were added to a solution of 5.03 x 10-2 M NaCl. A platinum electrode immersed in the layer of mercury was connected to a reference half-cell whose electrode potential was 0.222V. The voltage of the cell was 0.121 V with the reference electrode acting as the anode. Calculate the Ksp for Hg2Cl2. ( E0 for the required half reaction = 0.797 V)
To calculate the solubility product constant (Ksp) for Hg2Cl2, we need to use the Nernst equation and the relationship between the cell potential and the standard electrode potential.
First, let's write the balanced equation for the dissolution of Hg2Cl2:
Hg2Cl2 ⇌ 2Hg + 2Cl-
The Nernst equation is given by:
E = E0 - (RT/nF) * ln(Q)
Where:
- E is the cell potential (0.121 V in this case).
- E0 is the standard electrode potential (0.797 V in this case).
- R is the ideal gas constant (8.314 J/(mol·K)).
- T is the temperature in Kelvin (we'll assume it's 298 K).
- n is the number of electrons transferred (2 in this case since 2 Hg atoms are formed).
- F is the Faraday constant (96485 C/mol).
- Q is the reaction quotient, which is the ratio of the product concentrations raised to their stoichiometric coefficients.
We can rearrange the Nernst equation to solve for ln(Q):
ln(Q) = (E0 - E) * (nF/RT)
Now, to calculate Q, we need to find the concentrations of Hg2+ and Cl- ions. Since Hg2Cl2 dissolves into 2Hg and 2Cl-, the concentrations of Hg2+ and Cl- will be twice the concentration of Hg2Cl2.
Given:
Initial concentration of Hg2Cl2 = small amount
Volume of the solution = 25 ml = 0.025 L
Final concentration of Hg2Cl2 = 0 L (since it's completely precipitated)
Now, let's calculate the concentrations:
Initial moles of Hg2Cl2 = initial concentration * volume = small amount * 0.025 L
Final moles of Hg2Cl2 = 0 moles (it's completely precipitated)
Since the Hg2Cl2 dissolves into 2 Hg ions:
Initial moles of Hg = 2 * initial moles of Hg2Cl2
Final moles of Hg = 2 * final moles of Hg2Cl2 = 0 moles
Since the Hg2Cl2 dissolves into 2 Cl- ions:
Initial moles of Cl- = 2 * initial moles of Hg2Cl2
Final moles of Cl- = 2 * final moles of Hg2Cl2 = 0 moles
Therefore, the concentrations of Hg2+ and Cl- are both zero.
Now, substitute the known values into the equation:
ln(Q) = (0.797 V - 0.121 V) * (2 * 96485 C/mol) / (8.314 J/(mol·K) * 298 K)
Calculate ln(Q) using the above equation.
Finally, take the antilog of ln(Q) to get Q:
Q = e^(ln(Q))
Given that Ksp = [Hg2+]^2 * [Cl-]^2, substitute the values of [Hg2+] and [Cl-] (both are zero) into the equation to get Ksp = 0.
Therefore, the Ksp for Hg2Cl2 is 0.