The equilibrium constant of the reaction shown below is 2.41 × 108 at 25°C:

3Ag(s) + NO3—(aq) +4H+(aq) → 3Ag+(aq) + NO(g) + 2H2O(l)

Calculate the value of E°cell for a cell utilizing this reaction.

Question

The equilibrium constant of the reaction shown below is 2.41 × 108 at 25°C:

3Ag(s) + NO3—(aq) +4H+(aq) → 3Ag+(aq) + NO(g) + 2H2O(l)

Calculate the value of E°cell for a cell utilizing this reaction.

BlackTom AI · Accepted Answer

First, identify the number of electrons exchanged in the balanced redox reaction. The oxidation half-reaction is Ag(s) → Ag+(aq) + e−, which shows 1 electron per silver atom. With 3 Ag(s) involved, the total electrons transferred is n = 3. The reduction half-reaction is NO3− + 4 H+ + 3 e− → NO(g) + 2 H2O(l), which confirms the overall electron count matches at n = 3.

Now use the relation between the standard cell potential, E°cell, and the equilibrium constant, K, at 25°C for a reaction with n electrons: E°cell = (0.05916 V / n) × log10(K). Here n = 3 and K = 2.41 × 10^8.

Compute log10(K): log10(2.41 × 10^8) = log10(2.41) + log10(10^8) = 0.3820... + 8 = 8.3820 (approximately).

Therefore, E°cell = (0.05916 V / 3) × 8.3820 ≈ 0.01972 V × 8.3820 ≈ 0.165 V (to three sig figs).

Evaluating the given options:
- Option “–1.43 V”: This would require a large negative potential; it does not align with the calculated positive E°cell. The sign and magnitude are both inconsistent with the log(K) scaling for this n and K, so this is incorrect.
- Option “+0.0826 V”: This value is roughly half of the computed 0.165 V, which could arise if n were 6 (or if a miscalculation occurred with the 0.05916 constant). Since n = 3 here, this option does not fit the formula, so it is incorrect.
- Option “–2.41 V”: A large negative potential far from the calculated 0.165 V, and the sign is opposite of what the log(K) calculation yields for this reaction. This is incorrect.
- Option “+0.0718 V”: This is a smaller positive value that might come from an approximate calculation or rounding error, but it is not the correct result when using n = 3 and K = 2.41 × 10^8 with the standard constants.
- Option “+0.165 V”: This aligns precisely with the calculation E°cell ≈ 0.165 V for n = 3 and K = 2.41 × 10^8, using E° = (0.05916/n) log10(K). The magnitude and sign match the expected outcome.

In summary, the reasoning shows how the electron count and the K value determine E°cell, and among the options the one matching the calculated value is the positive 0.165 V. The other choices deviate in sign, magnitude, or come from incorrect assumptions about n or the relationship between E° and K.

The equilibrium constant of the reaction shown below is 2.41 × 108 at 25°C: 3Ag(s) + NO3—(aq) +4H+(aq) → 3Ag+(aq) + NO(g) + 2H2O(l) Calculate the value of E°cell for a cell utilizing this reaction.

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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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