Determine the equilibrium constant at 25°C for the reaction Zn + Ni²⁺ $\rightarrow$ Zn²⁺ + Ni
Zn²⁺ + 2e^- $\rightarrow$ Zn
E° = -0.76 V
Ni^-2+ + 2e^- $\rightarrow$ Ni
E° = -0.23 V

Question

Determine the equilibrium constant at 25°C for the reaction Zn + Ni²⁺ $\rightarrow$ Zn²⁺ + Ni Zn²⁺ + 2e^- $\rightarrow$Zn E° = -0.76 V Ni^-2+ + 2e^- $\rightarrow$ Ni E° = -0.23 V

Quizplus · Accepted Answer

The equilibrium constant (K) can be calculated using the formula:

K = [ZnS][Ni]/[NiS][Zn]

where the square brackets indicate concentration.

Since the stoichiometric coefficients of Zn and Ni are both 1, their concentrations can be assumed to be equal and constant at "x". Also, since the stoichiometric coefficients of the sulfides are both 1, their concentrations can also be assumed to be equal and constant at "y".

Therefore, the expression for K simplifies to:

K = x^2/y

The standard reduction potentials for the half-reactions can be used to calculate the standard cell potential (E°) for the overall reaction:

E°cell = E°Ni - E°Zn = -0.23 V - (-0.76 V) = 0.53 V

The Nernst equation can then be used to calculate the cell potential (E) at non-standard conditions:

E = E°cell - (RT/nF) ln(Q)

where R is the gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred in the balanced equation (in this case, n = 2), F is the Faraday constant, and Q is the reaction quotient.

At equilibrium, Q = K, so:

E = E°cell - (RT/nF) ln(K)

At 25°C (298 K), the equation becomes:

E = 0.53 V - (0.0257/2) ln(K)
K = e^(2(0.53 - E)/(0.0257))

Using the given standard reduction potentials, the equilibrium constant can be calculated as:

K = 8.6 x 10^10

Therefore, the best choice is C, 8.6 x 10^10.

Determine the Equilibrium Constant at 25°C for the Reaction Zn →\rightarrow→

Determine the Equilibrium Constant at 25°C for the Reaction Zn $\rightarrow$