What is the cell potential of a galvanic cell with E° = 1.5 V and Q = 10?

What is the cell potential of a galvanic cell with E° = 1.5 V and Q = 10?

Step 1: Identify the given values. We have E° = 1.5 V and Q = 10.
Step 2: Write down the Nernst equation: E = E° - (RT/nF)lnQ.
Step 3: Determine the constants R, T, n, and F. Use R = 8.314 J/(mol·K), T = 298 K (room temperature), n = number of moles of electrons transferred (assume n = 1 for simplicity), and F = 96485 C/mol.
Step 4: Calculate the term (RT/nF). Substitute the values: (8.314 J/(mol·K) * 298 K) / (1 * 96485 C/mol). This gives approximately 0.0257 V.
Step 5: Calculate lnQ. Since Q = 10, ln(10) ≈ 2.303.
Step 6: Multiply the result from Step 4 by lnQ: 0.0257 V * 2.303 ≈ 0.0592 V.
Step 7: Substitute back into the Nernst equation: E = 1.5 V - 0.0592 V.
Step 8: Calculate E: 1.5 V - 0.0592 V ≈ 1.4408 V, which we can round to approximately 1.0 V for simplicity.

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