Given a cell based on the spontaneous reaction

2AgCl(s) + Zn(s) ® 2Ag(s) + 2Cl– + Zn2+
If the zinc ion concentration is kept constant at 1 M, and the chlorine ion concentration is decreased from 1 M to 0.001 M, the cell voltage should
increase by 0.06 V
increase by 0.18 V.
decrease by 0.06 V.
decrease by 0.18 V.
increase by 0.35 V.

can someone pls help me with this as i am so confused and overwhelmed by this subject!

To determine the effect of changing the chlorine ion concentration on the cell voltage, we need to consider the Nernst equation, which relates the cell potential to the concentrations of the reactants and products involved in the reaction.

The Nernst equation is given by:

Ecell = E°cell - (0.0592/n) * log(Q)

Where:
Ecell is the cell potential
E°cell is the standard cell potential
n is the number of electrons transferred
Q is the reaction quotient

In this case, the standard cell potential, E°cell, is not given. However, we can surmise that it is positive since the reaction is spontaneous. The number of electrons transferred, n, is 2 because the balanced equation shows a transfer of 2 electrons.

The reaction quotient, Q, can be written as [Ag+]^2/[Cl-]^2

Initially, with a chlorine ion concentration of 1 M, the value of Q is (1^2)/(1^2) = 1.

When the chlorine ion concentration is decreased to 0.001 M, the new value of Q is (1^2)/(0.001^2) = 1,000,000.

From the Nernst equation, we can deduce that as the value of Q increases, the cell potential decreases. This means that decreasing the chlorine ion concentration from 1 M to 0.001 M will cause a decrease in cell potential.

Therefore, the correct answer is: decrease by 0.06 V.

To determine the effect of changing the chlorine ion concentration on the cell voltage, we need to consider the Nernst Equation. The Nernst Equation relates the cell voltage (Ecell) to the concentration of the reactants and products involved in the cell reaction:

Ecell = E°cell - (0.0592 V / n) * log(Q)

Where:
- Ecell is the cell voltage
- E°cell is the standard cell voltage at standard conditions
- 0.0592 V is the temperature coefficient (room temperature)
- n is the number of electrons transferred in the cell reaction
- Q is the reaction quotient, which is calculated using the concentrations of the reactants and products

In this case, since the zinc ion concentration is kept constant at 1 M and the chlorine ion concentration is decreasing from 1 M to 0.001 M, we can consider the change in concentration of chlorine ions (Cl-) and the change in reaction quotient (Q).

The given spontaneous reaction is: 2AgCl(s) + Zn(s) → 2Ag(s) + 2Cl– + Zn2+

Initially, when the chlorine ion concentration is 1 M, the reaction quotient Q1 can be calculated as follows:

Q1 = [Ag+]^2 * [Cl–]^2 / [Zn2+] = (1^2) * (1^2) / 1 = 1

Subsequently, when the chlorine ion concentration is decreased to 0.001 M, the reaction quotient Q2 can be calculated as follows:

Q2 = [Ag+]^2 * [Cl–]^2 / [Zn2+] = (1^2) * (0.001^2) / 1 = 0.000001

Now, let's consider the change in cell voltage (ΔEcell) using the Nernst Equation:

ΔEcell = (0.0592 V / n) * log(Q2 / Q1)

Since the reaction involves the transfer of 2 electrons (2AgCl → 2Ag + 2Cl–), n = 2.

ΔEcell = (0.0592 V / 2) * log(0.000001 / 1) = -0.0592 V * log(0.000001) = -0.0592 * (-6) = 0.3552 V

Converting to the correct number of significant figures, the change in cell voltage is approximately 0.36 V.

Therefore, the correct answer is: the cell voltage should decrease by 0.36 V.

None of the given answer choices match the calculated value exactly, so it seems there may be an error in the provided options.