Question

A concentration cell similar to the one shown is composed of two Cr electrodes and solutions of different Cr³⁺concentrations. The left compartment contains 0.409 M Cr³⁺, and the right compartment contains 0.834M Cr³⁺.  


1. Calculate the cell potential for this reaction at 298 K.

________volts

In this chromium concentration cell, the compartment on the left is the _________anode/cathode, and the compartment on the right is the _________anode/cathode.

2.A concentration cell similar to the one shown is composed of two Sn electrodes and solutions of different Sn²⁺concentrations. The left compartment contains 1.03 M Sn²⁺, and the right compartment contains 1.20 M Sn²⁺.  

Calculate the cell potential for this reaction at 298 K.

_______ volts

In this tin concentration cell, the reaction would proceed spontaneously _________from the left to the right compartment/from the right to the left compartment.

Answer 1

1. To calculate the cell potential for the chromium concentration cell, we can use the Nernst equation:

Ecell = E°cell - (RT / nF) * ln(Q)

Given: Concentration in the left compartment (Cr3+): [Cr3+]left = 0.409 M Concentration in the right compartment (Cr3+): [Cr3+]right = 0.834 M Temperature: T = 298 K

The reaction occurring in the cell can be written as:

Cr3+(left) + 3e- → Cr(s) Cr3+(right) + 3e- → Cr(s)

Since the same species (Cr) is being reduced at both electrodes, the E°cell value is 0.

The Nernst equation simplifies to:

Ecell = - (RT / nF) * ln(Q)

where R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, n is the number of electrons transferred in the balanced half-reaction (in this case, 3), F is the Faraday constant (96485 C/mol), and Q is the reaction quotient.

Q = ([Cr3+]right / [Cr3+]left)^3

Substituting the given values:

Q = (0.834 M / 0.409 M)^3

Calculating Q:

Q ≈ 6.587

Now, we can calculate the cell potential:

Ecell = - (RT / nF) * ln(Q)

Ecell = - ((8.314 J/(mol·K)) * (298 K) / (3 * 96485 C/mol)) * ln(6.587)

Ecell ≈ - (0.0257 V) * ln(6.587)

Ecell ≈ - 0.0257 V * 1.885

Ecell ≈ - 0.0484 V

Therefore, the cell potential for this reaction at 298 K is approximately -0.0484 volts.

In the chromium concentration cell, the compartment on the left is the cathode, and the compartment on the right is the anode.

2. To calculate the cell potential for the tin concentration cell, we can use the same process as above.

Given: Concentration in the left compartment (Sn2+): [Sn2+]left = 1.03 M Concentration in the right compartment (Sn2+): [Sn2+]right = 1.20 M Temperature: T = 298 K

The reaction occurring in the cell can be written as:

Sn2+(left) + 2e- → Sn(s) Sn2+(right) + 2e- → Sn(s)

Again, since the same species (Sn) is being reduced at both electrodes, the E°cell value is 0.

Using the Nernst equation, we can calculate the cell potential:

Ecell = - (RT / nF) * ln(Q)

where R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, n is the number of electrons transferred in the balanced half-reaction (in this case, 2), F is the Faraday constant (96485 C/mol), and Q is the reaction quotient.

Q = ([Sn2+]right / [Sn2+]left)^2

Substituting the given values:

Q = (1.20 M / 1.03 M)^2

Calculating Q:

Q ≈ 1.278

Now, we can calculate the cell potential:

Ecell = - (RT / nF) * ln(Q)

Ecell = - ((8.314 J/(mol·K)) * (298 K) / (2 * 96485 C/mol)) * ln(1.278)

Ecell ≈ - (0.0257 V) * ln(1.278)

Ecell ≈ - 0.0257 V * 0.245

Ecell ≈ - 0.0063 V

Therefore, the cell potential for this reaction at 298 K is approximately -0.0063 volts.

In the tin concentration cell, the reaction would proceed spontaneously from the left to the right compartment.