Electrochemistry

Step 1: Understanding the ionic activity coefficient.
The mean ionic activity coefficient is a measure of the effective concentration of ions in solution. It is defined as the geometric mean of the activity coefficients of the cations and anions involved. For a compound like , which dissociates into one ion and two ions, the mean ionic activity coefficient is given by the formula:

Step 2: Analyzing the options.
(A) : Incorrect — This does not account for the correct number of ions and their corresponding coefficients.
(B) : Incorrect — This does not represent the correct formula for the mean ionic activity coefficient.
(C) : Incorrect — This is not the correct formula for the mean ionic activity coefficient.
(D) : Correct — This is the correct formula for the mean ionic activity coefficient in this case.

Step 3: Conclusion.
The correct answer is (D) as it accurately represents the mean ionic activity coefficient for the dissociation of .

This is a thermodynamics problem involving the relationship between the electromotive force ( ) of an electrochemical cell, the temperature coefficient, and the standard enthalpy change ( ) of the cell reaction.

The relationship between the standard Gibbs free energy change ( ), the standard enthalpy change ( ), the temperature ( ), and the temperature coefficient of the is given by the Gibbs-Helmholtz equation for electrochemical cells:

$ \Delta G^\circ \text{emf} E n F \Delta G^\circ E \text{emf} \frac{dE}{dT} \text{emf} E 1.02 \text{ V} T 300 \text{ K} \frac{dE}{dT} -5.0 \times 10^{-5} \text{ V K}^{-1} F 96500 \text{ C mol}^{-1} n n=2 \Delta H^\circ \text{J mol}^{-1} \Delta H^\circ \Delta H^\circ 1 \text{ J} = 1 \text{ C} \cdot 1 \text{ V} \text{kJ mol}^{-1} 1000 \Delta H^\circ = -199.755 \text{ kJ mol}^{-1} $

Given: 0.01 M aqueous solution of Ca₃(PO₄)₂

Dissociation: $ $

Answer: 0.066 (rounded to three decimal places)

Given reaction: CuSO₄(aq) + Zn(s) → ZnSO₄(aq) + Cu(s)

Half-reactions:

• Reduction: Cu²⁺(aq) + 2e⁻ → Cu(s), E° = +0.34 V
• Oxidation: Zn(s) → Zn²⁺(aq) + 2e⁻, E° = +0.76 V (reverse of given)

Cell potential:

Relationship between ΔG° and E°: $ $

Answer: -212 kJ mol⁻¹ (rounded to nearest integer)

To find the total charge drawn during the process, we use the formula:
, where is the current in Amperes and is the time in seconds.
Given A and h, we need to convert the time from hours to seconds. There are 3600 seconds in an hour, so s.
Substitute these values into the formula:
.
The calculated charge is

To solve for the standard Gibbs free energy change (ΔG°) of the reaction in the galvanic cell, we start by acknowledging the reaction in question: Zn + Cu²⁺ → Zn²⁺ + Cu. The cell potential (E°_cell) is given as 1.10 V. The standard Gibbs free energy change is calculated using the formula:

ΔG° = -nFE°_cell

where n is the number of moles of electrons transferred in the balanced equation, F is Faraday's constant (96485 C mol^-1), and E°_cell is the standard cell potential (1.10 V).

In the given reaction, the oxidation and reduction half-reactions are:
Zn → Zn²⁺ + 2e^- (oxidation)
Cu²⁺ + 2e^- → Cu (reduction)

Here, n = 2 moles of electrons are transferred per mole of Zn reacted. Substituting these into the Gibbs free energy equation, we get:

ΔG° = -(2 mol)(96485 C/mol)(1.10 V)

Calculating this gives:
ΔG° = -212267 J (or -212.267 kJ, as 1 J = 0.001 kJ).

Rounding this value to the nearest integer, we obtain ΔG° = -212 kJ.

The Nernst equation for the standard EMF of the cell is given by:

Given the activities of the solids are unity and the activity of P⁺ is 2M, and the activity of Q⁺ is 1M, the standard EMF simplifies as follows:

• The mobility of ions in water is influenced by their size and hydration energy. Smaller ions tend to have higher mobility because they can move more easily through the solvent.
• Li⁺ is the smallest ion in this group and has the highest hydration energy, which reduces its mobility.
• K⁺ has the largest ionic radius and, thus, the lowest mobility among these ions.

Thus, the correct order of mobility is:

Therefore, the correct answer is (D)

We are given the following data:

• Mass of metal Q formed = 40 g
• Molar mass of Q = 40 g/mol (assumed)
• The gas constant R = 0.082 L atm mol⁻¹ K⁻¹
• Temperature = 298 K
• Pressure = 1 atm

First, we calculate the number of moles of Q formed:

Moles of Q =

Moles of Q =

Since one mole of Q produces one mole of Cl₂, we can now calculate the volume of Cl₂ using the ideal gas law:

PV = nRT

Substituting the known values:

V =

V = 12.1 L

Thus, the volume of Cl₂ formed is 12.1 litres.