How much will the potential of Zn2/Zn2+ change if the solution of Zn2+ is diluted 10 times?

To understand how the potential of the Zn²⁺/Zn half-cell changes when the Zn²⁺ solution is diluted 10 times, we can refer to the Nernst equation. This equation helps us calculate the cell potential based on the concentrations of the reactants and products involved in the electrochemical reaction.

The Nernst Equation Explained

The Nernst equation is given by:

E = E_° - (RT/nF) * ln(Q)

Where:

• E = cell potential under non-standard conditions
• E_° = standard cell potential
• R = universal gas constant (8.314 J/(mol·K))
• T = temperature in Kelvin
• n = number of moles of electrons transferred in the reaction
• F = Faraday's constant (96485 C/mol)
• Q = reaction quotient

Understanding the Reaction Quotient (Q)

For the Zn²⁺/Zn half-reaction:

Zn^2+- ⇌ Zn

The reaction quotient, Q, is defined as:

Q = [Zn²⁺]/[Zn]

Since the concentration of solid Zn does not change, we can simplify Q to just the concentration of Zn²⁺.

Effect of Dilution on Potential

When you dilute the Zn²⁺ solution by a factor of 10, the concentration of Zn²⁺ decreases. If we denote the initial concentration as [Zn²⁺]_initial and the diluted concentration as [Zn²⁺]_diluted, we have:

[Zn²⁺]_diluted = [Zn²⁺]_initial / 10

Calculating the Change in Potential

Substituting the new concentration into the Nernst equation, we can see how E changes:

E_diluted = E_° - (RT/nF) * ln([Zn²⁺]_diluted)

Since [Zn²⁺]_diluted is one-tenth of the initial concentration, the natural logarithm term will change significantly:

ln([Zn²⁺]_diluted) = ln([Zn²⁺]_initial / 10) = ln([Zn²⁺]_initial) - ln(10)

Thus, the potential will decrease because the logarithm of a smaller number is negative, which means:

E_diluted < E_initial

Conclusion on Potential Change

In summary, diluting the Zn²⁺ solution by a factor of 10 will result in a decrease in the potential of the Zn²⁺/Zn half-cell. The exact change in potential can be calculated if the initial concentration and temperature are known, but qualitatively, we can say that the potential will drop as the concentration of Zn²⁺ decreases.