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Grade 12th passGeneral Physics

How much potential of Zn/Zn2+ change if the solution of Zn2+ is diluted 1000 times?

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9 Years agoGrade 12th pass
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1 Answer

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ApprovedApproved Tutor Answer1 Year ago

To understand how the potential of the Zn/Zn2+ half-cell changes when the Zn2+ solution is diluted 1000 times, we need to consider the Nernst equation. This equation relates the concentration of ions in solution to the electrode potential. Let's break this down step by step.

The Nernst Equation

The Nernst equation is given by:

E = E0 - (RT/nF) * ln(Q)

Where:

  • E = electrode potential
  • E0 = standard electrode potential
  • R = universal gas constant (8.314 J/(mol·K))
  • T = temperature in Kelvin
  • n = number of moles of electrons transferred in the half-reaction
  • F = Faraday's constant (96485 C/mol)
  • Q = reaction quotient

Understanding the Reaction Quotient (Q)

For the zinc half-cell reaction:

Zn(s) ⇌ Zn2+(aq) + 2e-

The reaction quotient, Q, can be expressed as:

Q = [Zn2+]

When the solution is diluted 1000 times, the concentration of Zn2+ ions decreases significantly. If we assume the initial concentration of Zn2+ was 1 M, after dilution, it becomes 0.001 M.

Calculating the Change in Potential

Let’s plug the new concentration into the Nernst equation. Assuming standard conditions (25°C or 298 K), we can calculate the change in potential:

First, we need to determine the standard electrode potential for the Zn/Zn2+ couple, which is approximately -0.76 V.

Now, substituting the values into the Nernst equation:

E = -0.76 V - (0.0257 V/2) * ln(0.001)

Here, 0.0257 V is derived from (RT/F) at 298 K, and the factor of 2 comes from the number of electrons transferred in the half-reaction.

Evaluating the Logarithmic Term

Calculating the logarithmic term:

ln(0.001) = -6.907

Now substituting this back into the equation:

E = -0.76 V - (0.0257 V/2) * (-6.907)

E = -0.76 V + 0.0885 V

E ≈ -0.6715 V

Conclusion on Potential Change

After diluting the Zn2+ solution 1000 times, the potential of the Zn/Zn2+ half-cell increases from approximately -0.76 V to about -0.6715 V. This demonstrates that as the concentration of Zn2+ ions decreases, the electrode potential becomes less negative, indicating a greater tendency for the reduction of Zn2+ ions back to solid zinc. This principle is crucial in electrochemistry, as it highlights how concentration changes can significantly affect electrode behavior.