In a concentration cell, if the concentration of the anode is 0.1 M and the cath

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Question: In a concentration cell, if the concentration of the anode is 0.1 M and the cathode is 1 M, what is the cell potential at 25°C?

Options:

Correct Answer: 0.118 V

Solution:

E = (0.059 V/n) * log([Cathode]/[Anode]) = (0.059 V/2) * log(1/0.1) = 0.059 V * 1 = 0.059 V.

In a concentration cell, if the concentration of the anode is 0.1 M and the cathode is 1 M, what is the cell potential at 25°C?

In a concentration cell, if the concentration of the anode is 0.1 M and the cathode is 1 M, what is the cell potential at 25°C?

Step 1: Identify the concentrations of the anode and cathode. The anode concentration is 0.1 M and the cathode concentration is 1 M.
Step 2: Determine the number of electrons transferred in the reaction. For this example, we will assume n = 2.
Step 3: Use the formula for cell potential: E = (0.059 V/n) * log([Cathode]/[Anode]).
Step 4: Substitute the values into the formula: E = (0.059 V/2) * log(1/0.1).
Step 5: Calculate log(1/0.1). Since 1/0.1 = 10, log(10) = 1.
Step 6: Now substitute log(10) back into the equation: E = (0.059 V/2) * 1.
Step 7: Calculate (0.059 V/2) which equals 0.0295 V.
Step 8: Therefore, the cell potential E = 0.0295 V.

Nernst Equation – The Nernst equation relates the cell potential to the concentrations of the reactants and products in an electrochemical cell.
Concentration Cells – A concentration cell generates voltage from the difference in concentration of the same species at two electrodes.
Logarithmic Calculations – Understanding how to manipulate logarithmic expressions is crucial for calculating cell potential.