Under non-standard conditions cell potential (Ecell) may be calculated using the Nernst equation: The following cell reaction has an E0cell potential of 0.83 Volts: A(s) + B+(aq) → A+(aq) + B(s) Calculate the cell potential (Ecell) at 298 K (in Volts) if the concentration of A+ = 0.050 mol/L and the concentration of B+ = 5.0 mol/ L. Assume standard conditions. Give your answer to two decimal places. Do not include units. Cell potential =

Question

Under non-standard conditions cell potential (Ecell) may be calculated using the Nernst equation: The following cell reaction has an E0cell potential of 0.83 Volts: A(s) +  B+(aq) →  A+(aq) +  B(s) Calculate the cell potential (Ecell) at 298 K (in Volts) if the concentration of A+ = 0.050 mol/L and the concentration of B+ = 5.0 mol/ L. Assume standard conditions. Give your answer to two decimal places. Do not include units. Cell potential =

BlackTom AI · Accepted Answer

Start by identifying the given information and the formula to use. The problem provides E°cell = 0.83 V, T = 298 K, n = 1, and the reaction A(s) + B+(aq) → A+(aq) + B(s). The Nernst equation in the form given by the image uses the natural logarithm: Ecell = E°cell − (RT / (nF)) ln(Q). We need to determine Q from the concentrations of the aqueous species involved. In this reaction, solids A and B are omitted from the reaction quotient, so Q = [A+]/[B+]. With [A+] = 0.050 M and [B+] = 5.0 M, Q = 0.050 / 5.0 = 0.01.

Next, compute the constant RT/(nF) at 298 K: R = 8.314 J·mol−1·K−1, T = 298 K, F = 96485 C·mol−1, n = 1. So RT/(nF) = (8.314 × 298) / 96485 ≈ 2477.572 / 96485 ≈ 0.02569 (V per ln unit).

Compute ln(Q): Q = 0.01, so ln(Q) = ln(0.01) = ln(1/100) = −ln(100) ≈ −4.605.

Now substitute into the Nernst equation:
Ecell = E°cell − (RT/(nF)) ln(Q) = 0.83 − (0.02569)(−4.605) = 0.83 + 0.1183 ≈ 0.9483 V.
Rounding to two decimal places as requested yields Ecell ≈ 0.95.

Why this matches the provided answer: because the equation shown uses the natural logarithm (ln) rather than log base 10, which increases the magnitude of the correction for small Q values, leading to a larger Ecell than if log10 were used.

Why the other possible interpretations would be wrong:
- If you used log base 10 instead of ln (log10 Q), the correction term would be (RT/nF) log10(Q) with log10(Q) = log10(0.01) = −2, giving a smaller correction of about 0.0514, and Ecell ≈ 0.881 V, not 0.95 V.
- If you miscomputed Q as 0.050/0.505 or any other ratio, you would obtain an incorrect Q value, leading to a wrong Ecell. Proper Q must reflect only the aqueous species in the reaction quotient and exclude solids.
- If you forgot the negative sign in the ln(Q) term, you would incorrectly subtract a negative amount, altering the final value.

Therefore, the correct Ecell at 298 K with the given concentrations is approximately 0.95 (dimensionless here, as requested no units).

类似问题

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A voltaic cell is constructed using a Hg22+/Hg couple (E° = +0.85 V) and a Ag+/Ag couple (E° = +0.80 V) as the electrodes. Calculate [Ag+] if [Hg22+] = 0.24 M and Ecell = +0.17 V at 25 °C.

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