The equilibrium constant for the reaction,

Ni(s) + Hg2Cl2(s) → 2Hg(l) + 2Cl–(aq) + Ni2+(aq)

is 1.8
×
1019 at 25 °C. Calculate the value of E°cell for a cell utilizing this reaction.

Question

The equilibrium constant for the reaction,

Ni(s) + Hg2Cl2(s) → 2Hg(l) + 2Cl–(aq) + Ni2+(aq)

is 1.8 
×
 1019 at 25 °C. Calculate the value of E°cell for a cell utilizing this reaction.

BlackTom AI · Accepted Answer

The problem asks to calculate the standard cell potential E°cell for the given redox reaction using the equilibrium constant K at 25 °C. The options are:
- Question: The equilibrium constant for the reaction Ni(s) + Hg2Cl2(s) → 2 Hg(l) + 2 Cl–(aq) + Ni2+(aq) is 1.8 × 10^19 at 25 °C. Calculate the value of E°cell for a cell utilizing this reaction.
- Answer options: 
  1) +0.57 V 
  2) –1.14 V 
  3) +1.14 V 
  4) –0.57 V

Option 1: +0.57 V. To relate K to E°, we use the relation E°cell = (RT/nF) ln K. At 25 °C, RT/F ≈ 0.025693 V, so RT/(nF) = 0.025693/n. The overall reaction involves Ni(0) being oxidized to Ni2+ (loss of 2 electrons) and Hg2Cl2 being reduced to Hg(0) (gain of 2 electrons via Hg2Cl2 + 2e– → 2 Hg + 2 Cl–). Thus, n = 2 electrons transferred. Computing E°cell:
E°cell = (0.025693/2) × ln(1.8 × 10^19).
First find ln(1.8 × 10^19) = ln(1.8) + 19 ln(10) ≈ 0.5878 + 19 × 2.302585 ≈ 0.5878 + 43.7491 ≈ 44.3369.
Then E°cell ≈ 0.0128465 × 44.3369 ≈ 0.5695 V, which rounds to about +0.57 V. This is the value you’d expect for a large positive K, reflecting a strongly favorable reaction in the written direction.

Option 2: –1.14 V. A negative E°cell would imply a non-spontaneous forward reaction under standard conditions, which contradicts the extremely large positive K given (1.8 × 10^19). Since a large K corresponds to a favorable tendency for products at standard state, a negative E°cell here is inconsistent with the thermodynamics.

Option 3: +1.14 V. While a positive E°cell is plausible for a favorable reaction, the magnitude here is roughly double what the calculation from K yields. The relation E°cell = (RT/nF) ln K with n = 2 and K = 1.8 × 10^19 gives ~0.57 V, not ~1.14 V, so this overestimates according to the given K value.

Option 4: –0.57 V. This option has the wrong sign, implying the reaction as written would be non-spontaneous under standard conditions. Given the large K value, the forward reaction is strongly favored, so a negative E°cell does not fit the thermodynamic data.

In summary, the calculation using the standard thermodynamic relationship between the equilibrium constant and E°cell, with n = 2 and K = 1.8 × 10^19 at 25 °C, yields a value near +0.57 V, aligning with the first option.

The equilibrium constant for the reaction, Ni(s) + Hg2Cl2(s) → 2Hg(l) + 2Cl–(aq) + Ni2+(aq) is 1.8 × 1019 at 25 °C. Calculate the value of E°cell for a cell utilizing this reaction.单项选择题

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