Question

Consider the following reactions and standard electrode potentials at 25 ∘C:

Zn2+(aq)+2e− → Zn(s) E∘=−0.76 V

Fe2+(aq)+2e−→ Fe(s) E∘= −0.45 V

Compute the equilibrium constant at 25 ∘C for the reaction between Zn2+(aq) and Fe(s), which form Zn(s) and Fe2+(aq).

Answer 1

To calculate the equilibrium constant for the reaction between Zn2+(aq) and Fe(s), we need to use the standard electrode potentials of the two half-reactions. The standard cell potential (E∘ cell) can be calculated by subtracting the potentials of the two half-reactions. Using the Nernst equation, we can then relate E∘ cell to the equilibrium constant (K) by considering the reaction quotient (Q).

Step-by-step explanation:

To calculate the equilibrium constant for the reaction between Zn2+(aq) and Fe(s), we need to use the standard electrode potentials of the two half-reactions. The reduction half-reaction for Zn2+(aq) is Zn2+(aq) + 2e− → Zn(s) with an E∘ of -0.76 V, and the reduction half-reaction for Fe2+(aq) is Fe2+(aq) + 2e− → Fe(s) with an E∘ of -0.45 V. Since the overall reaction involves the reduction of Zn2+(aq) and the oxidation of Fe(s), we can subtract the two half-reaction potentials to calculate the standard cell potential. In this case, the standard cell potential (E∘ cell) would be -0.76 V - (-0.45 V) = -0.31 V.

Next, we can use the Nernst equation to relate the equilibrium constant (K) with the standard cell potential (E∘ cell). The Nernst equation is given by:

E = E∘ - (RT/nF)ln(Q)

Where E is the cell potential under nonstandard conditions, E∘ is the standard cell potential, R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, n is the number of electrons transferred in the reaction (in this case, 2), F is the Faraday constant (96,485 C/mol), and Q is the reaction quotient.

The reaction quotient (Q) can be calculated using the concentrations of the species involved in the reaction. In this case, it would be Q = [Zn2+(aq)] / ([Fe(s)]^2).

Using the calculated E∘ cell and the given values for the standard electrode potentials of the half-reactions, we can substitute them into the Nernst equation to solve for the equilibrium constant (K).

Answer 2

The equilibrium constant
`(\(K\))` for the reaction between
`Zn\(^(2+)\)` and Fe to form Zn and
`Fe\(^(2+)\)` at 25°C is extremely small, indicating that the reaction heavily favors the reactants.

To find the equilibrium constant
`(\(K\))` for the reaction between
`Zn\(^(2+)\)` and Fe, utilize the Nernst equation and the given standard electrode potentials.

The reaction you're interested in is:

`\[ \text{Zn}^(2+)(aq) + \text{Fe}(s) \rightarrow \text{Zn}(s) + \text{Fe}^(2+)(aq) \]`

The standard cell potential
`(\(E^\circ_{\text{cell}}\))` can be determined by subtracting the standard reduction potentials:

`\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]`

For the given reactions:

Cathode:
`Zn\(^(2+)\)(aq) + 2e\(^-\) → Zn(s) (with \(E^\circ = -0.76\) V)`

Anode:
`Fe\(^(2+)\)(aq) + 2e\(^-\) → Fe(s) (with \(E^\circ = -0.45\) V)`

Thus,

`\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (-0.76 \, \text{V}) - (-0.45 \, \text{V}) = -0.31 \, \text{V} \]`

Now, using the Nernst equation to relate the standard cell potential to the equilibrium constant
`(\(K\))`:

`\[ E_{\text{cell}} = E^\circ_{\text{cell}} - (RT)/(nF) \ln K \]`

Given that
`\(E_{\text{cell}} = -0.31`,
`\text{V}\) and \(E^\circ_{\text{cell}}\)` has already been calculated, and at 25°C
`(\(T = 298 \, \text{K}\))`, the number of electrons involved in the balanced equation
`(\(n = 2\))`, and
`\(F\)` is Faraday's constant
`(\(F = 96485 \, \text{C/mol}\))`.

Let's rearrange the equation to solve for
`\(K\)`:

`\[ K = e^{\left((nF)/(RT) \cdot (E^\circ_{\text{cell}} - E_{\text{cell}})\right)} \]`

Now plug in the values:

`\[ K = e^{\left(\frac{2 * 96485 \, \text{C/mol}}{8.314 \, \text{J/(mol} \cdot \text{K)}} \cdot \frac{-0.31 \, \text{V}}{298 \, \text{K}}\right)} \]`

`\[ K \approx e^{\left(\frac{2 * 96485 \, \text{C/mol}}{8.314 \, \text{J/(mol} \cdot \text{K)}} \cdot -1.04\right)} \]`

`\[ K \approx e^(-250.9) \]`

`\[ K \approx 2.34 * 10^(-109) \]`