A Zn electrode is immersed in a solution that is \(1.00 \mathrm{M}\) in \(\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) and $1.00 \mathrm{M}\( in \)\mathrm{NH}_{3}$. When the cathode is a standard hydrogen electrode, the emf of the cell is found to be \(+1.04 \mathrm{~V}\). What is the formation constant for \(\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+} ?\)

We need to identify the half-reactions and the overall cell reaction before finding the formation constant of [Zn(NH3)4]^2+. The half-reactions for this system are: (1) Zn^2+ + 2e^- ⟶ Zn (Reduction reaction at the Zn electrode) (2) 2H^+ + 2e^- ⟶ H2 (Reduction reaction at the standard hydrogen electrode, SHE) Now, we will write down the overall cell reaction by adding the two half-reactions: Zn^2+ + 2H^+ + 2e^- ⟶ Zn + H2 + 2e^- Zn^2+ + 2H^+ ⟶ Zn + H2

Now, we will use the Nernst equation to relate the cell potential (EMF) and the concentrations of the various species. The Nernst equation is: E = E° - (RT/nF) * ln(Q) Since the given cell potential is +1.04 V, we can find the equilibrium constant (K) using the relation between E, E°, and K: 1.04 V = 0 V - (RT/2F) * ln(Q) Here, E° = 0 V is the standard cell potential, R = 8.314 J/(mol K) is the gas constant, T = 298.15 K (assuming room temperature), F = 96485 C/mol is Faraday's constant, and n = 2 is the number of electrons transferred. Now, let's find the reaction quotient (Q) for the cell reaction: Q = [Zn^2+]/[H^+]^2 Since the solution is 1.00 M in NH3, we can set up an equilibrium expression for the formation constant Kf of the complex, relating Kf, the initial concentration of NH3 and the equilibrium concentration of Zn^2+: Kf = [Zn(NH3)4]^2+ / ([Zn^2+][NH3]^4) We have found the value of Q from Nernst equation above: Q = [Zn^2+]/[H^+]^2 Therefore, Kf = [Zn(NH3)4]^2+ / ([Zn^2+][NH3]^4) = [Zn^2+]/[H^+]^2 We will use this relationship to find the formation constant Kf after finding the concentration of Zn^2+. Solving the Nernst equation for [Zn^2+], we get: 1.04 V = - (RT/2F) * ln(([Zn^2+])/[H^+]^2) Rearranging the equation to isolate [Zn^2+]: [Zn^2+] = [H^+]^2 * exp(-2F(1.04 V)/(RT)) Using R = 8.314 J/(mol K), T = 298.15 K, and F = 96485 C/mol, we find: [Zn^2+] ≈ 2.92 × 10^-9 M

Now that we have the concentration of Zn^2+, we can calculate the formation constant Kf for the [Zn(NH3)4]^2+ complex. Using the relationship derived in step 2: Kf = [Zn(NH3)4]^2+/([Zn^2+][NH3]^4) = [Zn^2+]/[H^+]^2 Since the complex and NH3 have equal concentration 1.00 M given by the problem: Kf = [Zn^2+]/[H^+]^2 = (2.92 × 10^-9 M) / (1.00 M)^2 Kf ≈ 2.92 × 10^-9 Hence, the formation constant for the [Zn(NH3)4]^2+ complex is approximately 2.92 × 10^-9.