(a) When 300.0 milliliters of a solution of 0.200 molar AgNO3 is mixed with 100.0 milliliters of a 0.0500 molar CaCl2 solution, what is the concentration of silver ion after the reaction has gone to completion?

Question

(a) When 300.0 milliliters of a solution of 0.200 molar AgNO3 is mixed with 100.0 milliliters of a 0.0500 molar CaCl2 solution, what is the concentration of silver ion after the reaction has gone to completion?

(b)Write the net cell reaction for a cell formed by placing a silver electrode in the solution remaining from the reaction above and connecting it to a standard hydrogen electrode.

(c)Calculate the voltage of a cell of this type in which the concentration of silver ion is 410-2 M.

(d)Calculate the value of the standard free energy change ∆G˚ for the following half reaction:
Ag+ (1M) + e- ⁄ Ag˚

Answer 1

(a) Oh, chemistry, you really know how to mix things up! So, we have 300.0 ml of 0.200 M AgNO3 and 100.0 ml of 0.0500 M CaCl2. Let's see what happens when we put them together. mixes imaginary beakers Ta-da! After the reaction goes to completion, the concentration of silver ion will be... wait for it... drum roll ... the same as my chances of becoming a Nobel Prize-winning scientist... zero!

(b) Now, let's build ourselves an electric circus! Take the solution from the previous problem (where silver ion concentration was zero) and toss in a silver electrode. Connect it to a standard hydrogen electrode and voila! You've got yourself a cell. *places electrode in solution* The net cell reaction? Well, it's like watching a clown juggling hydrogen and silver ions: 2Ag+ (aq) + 2H2(g) → 2Ag(s) + H2(g). Quite the electrifying show!

(c) Ah, let's light up the stage with some voltage magic. In this cell, when the concentration of silver ion is 4x10^-2 M, the voltage will be... wait for it... *cue suspenseful music* ... I'm sorry, I don't know the answer! Guess I'll have to stick to telling jokes instead of calculating voltages.

(d) Time to flex those chemistry muscles! The value of the standard free energy change (∆G˚) for the half reaction Ag+ (1M) + e- ⁄ Ag˚ is like trying to calculate how many circus peanuts fit in a clown car. It's complex! But hey, who needs ∆G when you've got ∆G (delta grin) from my jokes? It's off the charts!

Answer 2

(a) To find the concentration of silver ion after the reaction has gone to completion, we need to determine the limiting reactant and use stoichiometry.

Step 1: Calculate the number of moles of AgNO3 and CaCl2:
Moles of AgNO3 = volume (L) x molarity = (300.0 mL / 1000 mL/L) x 0.200 mol/L
Moles of CaCl2 = volume (L) x molarity = (100.0 mL / 1000 mL/L) x 0.0500 mol/L

Step 2: Determine the limiting reactant:
The reactant that produces the least amount of product, based on its stoichiometry, is the limiting reactant. In this case, we compare the moles of AgNO3 and CaCl2.

The balanced chemical equation for the reaction is:
2AgNO3 + CaCl2 -> 2AgCl + Ca(NO3)2

From the balanced equation, we can see that 2 moles of AgNO3 react with 1 mole of CaCl2 to produce 2 moles of AgCl. Therefore, the ratio of moles of AgNO3 to CaCl2 is 2:1.

The moles of AgNO3: 300.0 mL x 0.200 mol/L = 0.0600 mol
The moles of CaCl2: 100.0 mL x 0.0500 mol/L = 0.00500 mol

Since the ratio of AgNO3 to CaCl2 is 2:1, we can see that CaCl2 is the limiting reactant because it produces the least amount of product.

Step 3: Calculate the moles of AgCl formed:
From the balanced equation, we know that 1 mole of CaCl2 produces 2 moles of AgCl. Therefore, the moles of AgCl formed is:

Moles of AgCl = (0.00500 mol CaCl2) x (2 mol AgCl / 1 mol CaCl2) = 0.0100 mol

Step 4: Calculate the concentration of silver ion:
The concentration of silver ion is equal to the moles of AgCl formed divided by the total volume of the solution.

Total volume of the solution = volume of AgNO3 + volume of CaCl2 = 300.0 mL + 100.0 mL = 400.0 mL = 0.4 L

Concentration of silver ion = (0.0100 mol AgCl) / (0.4 L) = 0.0250 M

Therefore, the concentration of silver ion after the reaction has gone to completion is 0.0250 M.

(b) The net cell reaction for a cell formed by placing a silver electrode in the solution remaining from the reaction above and connecting it to a standard hydrogen electrode can be determined by considering the Redox reactions involved.

The half-reactions involved in this case are:
Ag+ (in solution) + e- -> Ag (on the electrode) (reduction half-reaction)
2H+ (in solution) + 2e- -> H2 (gas on the electrode) (oxidation half-reaction)

The net cell reaction can be obtained by combining these half-reactions:
2Ag+ (in solution) + 2H+ (in solution) -> 2Ag (on the electrode) + H2 (gas on the electrode)

(c) To calculate the voltage of the cell when the concentration of silver ion is 4x10-2 M, we can use the Nernst equation:

Ecell = E°cell - (0.0592 V / n) x log10(Q)

Where:
Ecell is the cell potential
E°cell is the standard cell potential (from a table)
n is the number of moles of electrons transferred in the balanced equation
Q is the reaction quotient, which is the ratio of the concentrations of the products to reactants raised to their stoichiometric coefficients.

Looking at the net cell reaction:
2Ag+ (in solution) + 2H+ (in solution) -> 2Ag (on the electrode) + H2 (gas on the electrode)
The balanced equation shows that 2 mol of electrons are transferred.

Given:
[Ag+] = 4x10-2 M
[H+] = 1 M (assuming it is standard conditions)

Plugging the values into the Nernst equation:
Ecell = E°cell - (0.0592 V / 2) x log10((4x10-2 M)² / (1 M)²)

The value of E°cell can be determined from a table of standard reduction potentials.

(d) To calculate the value of the standard free energy change (∆G˚) for the half-reaction:
Ag+ (1M) + e- -> Ag˚
We can use the equation:

∆G˚ = -nFE°

Where:
∆G˚ is the standard free energy change
n is the number of moles of electrons transferred in the balanced equation (1 in this case)
F is the Faraday constant (96,487 C/mol)
E° is the standard cell potential, which can be determined from a table of standard reduction potentials.

Plugging in the values, we have:
∆G˚ = -(1 mol)(96,487 C/mol)(E°)

Answer 3