A galvanic cell is composed of a standard Zn electrode and a Chromium electrode immersed in a solution containing Cr3+ . At what concentration of Cr3+ will the emf of the cell be zero?

To determine the concentration of Cr³⁺ at which the electromotive force (emf) of a galvanic cell composed of a zinc electrode and a chromium electrode is zero, we can apply the Nernst equation. This equation relates the concentration of ions in solution to the cell potential, allowing us to find the specific conditions under which the cell's emf becomes zero.

Understanding the Components of the Cell

In this galvanic cell, we have two half-reactions occurring at the electrodes:

• Anode (oxidation): Zn(s) → Zn²⁺(aq) + 2e^-
• Cathode (reduction): Cr³⁺(aq) + 3e^- → Cr(s)

Standard Electrode Potentials

The standard electrode potentials (E^°) for these reactions are crucial for our calculations:

• For zinc: E^°(Zn²⁺/Zn) = -0.76 V
• For chromium: E^°(Cr³⁺/Cr) = -0.74 V

Applying the Nernst Equation

The Nernst equation is given by:

E = E^° - (RT/nF) * ln(Q)

Where:

• E = cell potential
• 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

Calculating the Standard Cell Potential

First, we need to calculate the standard cell potential (E^°) for the galvanic cell:

E^° = E^°(cathode) - E^°(anode) = (-0.74 V) - (-0.76 V) = 0.02 V

Setting the Cell Potential to Zero

To find the concentration of Cr³⁺ that makes the emf zero, we set E = 0 in the Nernst equation:

0 = 0.02 V - (RT/nF) * ln(Q)

Rearranging gives:

ln(Q) = (nF/RT) * 0.02 V

Determining the Reaction Quotient (Q)

The reaction quotient Q for the cell can be expressed as:

Q = [Zn²⁺] / [Cr³⁺]

Assuming a standard concentration of Zn²⁺ is 1 M, we have:

Q = 1 / [Cr³⁺]

Substituting Values into the Equation

Now, substituting Q into our equation:

ln(1 / [Cr³⁺]) = (nF/RT) * 0.02 V

For this reaction, n = 3 (since 3 electrons are transferred in the reduction of Cr³⁺). Plugging in the values:

ln([Cr³⁺]) = - (nF/RT) * 0.02 V

Final Calculation

Using R = 8.314 J/(mol·K), T = 298 K (approximately room temperature), and F = 96485 C/mol:

ln([Cr³⁺]) = - (3 * 96485 C/mol / (8.314 J/(mol·K) * 298 K)) * 0.02 V

Calculating the right side gives us:

ln([Cr³⁺]) ≈ -0.24

Exponentiating both sides to solve for [Cr³⁺]:

[Cr³⁺] ≈ e^-0.24 ≈ 0.79 M

Thus, the concentration of Cr³⁺ at which the emf of the cell is zero is approximately 0.79 M. This means that when the concentration of chromium ions in the solution reaches this level, the driving force for the cell reaction is balanced, resulting in no net voltage output.