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Calculate the cell potential of the concentration cell described by Al(s)|Al³⁺(aq, 0.10 M)||Al³⁺(aq, 3.00 M)|Al(s) E°Al = -1.66 V, T = 25.0 °C

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Final Answer:

The cell potential of the concentration cell described by Al(s)|Al³⁺(aq, 0.10 M)||Al³⁺(aq, 3.00 M)|Al(s) at 25.0 °C can be calculated using the Nernst equation. The cell potential (Ecell) is 0.45 volts.

Step-by-step explanation:

In this concentration cell, the reduction half-reaction occurring at both electrodes involves Al³⁺ ions gaining electrons to form aluminum metal. The standard reduction potential (E°) for Al³⁺ to Al(s) is -1.66 volts. The Nernst equation relates the cell potential (Ecell) to the standard cell potential (E°cell), temperature (T), the gas constant (R), and the concentrations of species involved:

Ecell = E°cell - (RT/nF) * ln(Q)

Here, R is the gas constant (8.314 J/(mol*K)), T is the temperature in Kelvin (298 K), n is the number of electrons transferred (3 in this case), F is the Faraday constant (96485 C/mol), and Q is the reaction quotient.

The Nernst equation for this concentration cell becomes:

Ecell = E°cell - (0.0592/n) * log(Q)

Substituting the given values into the equation:

Ecell = -1.66 V - (0.0592/3) * log(0.10/3.00)

Ecell = -1.66 V + 0.0197 * log(0.0333)

Ecell = -1.66 V + 0.0197 * (-1.48)

Ecell = -1.66 V - 0.0291

Ecell = 0.45 V

The positive value of 0.45 volts indicates that the cell potential is positive, meaning the reaction is feasible in the given conditions. This calculation shows that the concentration gradient between the two solutions drives the cell to produce a potential of 0.45 volts.

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