If the concentration of Cu²⁺ in a cell is 0.01 M and the standard reduction pote
Practice Questions
Q1
If the concentration of Cu²⁺ in a cell is 0.01 M and the standard reduction potential is +0.34 V, what is the cell potential at 25°C using the Nernst equation?
0.34 V
0.30 V
0.28 V
0.25 V
Questions & Step-by-Step Solutions
If the concentration of Cu²⁺ in a cell is 0.01 M and the standard reduction potential is +0.34 V, what is the cell potential at 25°C using the Nernst equation?
Step 1: Identify the given values. The concentration of Cu²⁺ is 0.01 M, the standard reduction potential (E°) is +0.34 V, and the temperature is 25°C.
Step 2: Convert the temperature from Celsius to Kelvin. 25°C is equal to 298 K (25 + 273).
Step 3: Use the Nernst equation: E = E° - (RT/nF)ln(Q).
Step 4: Determine the values for R, T, n, and F. R (universal gas constant) = 0.0257 V·K/C, T = 298 K, n (number of electrons transferred) = 2 for Cu²⁺, and F (Faraday's constant) = 96485 C/mol.
Step 5: Calculate Q, the reaction quotient. Since the concentration of Cu²⁺ is 0.01 M, Q = 1 / [Cu²⁺] = 1 / 0.01 = 100.
Step 6: Substitute the values into the Nernst equation: E = 0.34 - (0.0257/2)ln(100).
Step 7: Calculate (0.0257/2) = 0.01285.
Step 8: Calculate ln(100) which is approximately 4.605.
Step 9: Multiply 0.01285 by 4.605 to get approximately 0.0592.
Step 10: Subtract this value from 0.34: E = 0.34 - 0.0592 = 0.2808 V.
Step 11: Round the final answer to two decimal places: E ≈ 0.30 V.
