Consider the concentration cell: b. Indicate the direction of electron flow.

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Identify the two half-cells involved in the concentration cell. A concentration cell consists of two half-cells with the same electrodes and electrolytes but different concentrations.

Determine which half-cell has the higher concentration of ions. In a concentration cell, electrons flow from the half-cell with lower ion concentration to the half-cell with higher ion concentration.

Use the Nernst equation to understand the potential difference between the two half-cells. The Nernst equation is E = E^0 - (RT/nF)ln(Q), where Q is the reaction quotient.

Recognize that the flow of electrons is from the anode to the cathode. In a concentration cell, the anode is the half-cell with the lower concentration, and the cathode is the half-cell with the higher concentration.

Conclude that the direction of electron flow is from the half-cell with lower concentration to the half-cell with higher concentration, driven by the concentration gradient.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Concentration Cell

A concentration cell is a type of electrochemical cell where two half-cells have the same electrodes but different concentrations of electrolyte solutions. The difference in concentration creates a potential difference, driving the flow of electrons from the anode (lower concentration) to the cathode (higher concentration). This setup illustrates how chemical potential can be converted into electrical energy.

Electron Flow Direction

In electrochemical cells, electrons flow from the anode to the cathode. In a concentration cell, the anode is the electrode in the half-cell with the lower concentration of ions, while the cathode has the higher concentration. This flow occurs because electrons move towards the region of higher potential energy, which corresponds to the higher concentration of reactants.

Nernst Equation

The Nernst equation relates the cell potential to the concentrations of the reactants and products in an electrochemical cell. It allows for the calculation of the voltage produced by a concentration cell based on the concentration difference. Understanding this equation is crucial for predicting the direction of electron flow and the overall behavior of the cell under varying conditions.

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The Nernst Equation

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Textbook Question

A voltaic cell consists of a Pb/Pb²⁺ half-cell and a Cu/Cu²⁺ half-cell at 25°C. The initial concentrations of Pb²⁺ and Cu²⁺ are 0.0500 M and 1.50 M, respectively. b. What is the cell potential when the concentration of Cu²⁺ has fallen to 0.200 M?

Textbook Question

A voltaic cell consists of a Pb/Pb²⁺ half-cell and a Cu/Cu²⁺ half-cell at 25°C. The initial concentrations of Pb²⁺ and Cu²⁺ are 0.0500 M and 1.50 M, respectively. c. What are the concentrations of Pb²⁺ and Cu²⁺ when the cell potential falls to 0.35 V?

Textbook Question

Make a sketch of a concentration cell employing two Zn/Zn²⁺ half-cells. The concentration of Zn²⁺ in one of the half-cells is 2.0 M and the concentration in the other half-cell is 1.0×10^–3 M. Label the anode and the cathode and indicate the half-reaction occuring at each electrode. Also indicate the direction of electron flow.

Textbook Question

Consider the concentration cell:

c. Indicate what happens to the concentration of Pb²⁺ in each half-cell.

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Textbook Question

A concentration cell consists of two Sn/Sn²⁺ half-cells. The cell has a potential of 0.10 V at 25°C. What is the ratio of the Sn²⁺ concentrations in the two half-cells?

Textbook Question

A Cu/Cu²⁺ concentration cell has a voltage of 0.22 V at 25 °C. The concentration of Cu²⁺ in one of the half-cells is 1.5×10^–3 M. What is the concentration of Cu²⁺ in the other half-cell? (Assume the concentration in the unknown cell is the lower of the two concentrations.)

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