The value of the equilibrium constant for the reaction of nickel(11) ions with cadmium metal is 1.17 x 105. Calculate AG° for the reaction at 25°C. Ni2+(aq) + Cd(s) Cd2*(aq) + Ni(s) O 5.07 kJ 28.9 kJ - 12.6 kJ - 28.9 kJ O 12.6 kJ

Calculate AG for the following reaction from the equilibrium constant at the temperature given. 2SO, (g) + Oz (g) --→ 2SO, (3) T=500°C Kp =48.2 O 24.9 kJ 0-24.9 kJ O 16.1 kJ 0 16.1 kJ

Calculate ΔG° for the following balanced redox reaction and determine if the reaction is spontaneous. (values are made-up)          2 Na+(aq) + 2 Cl-(aq) → 2 Na(s) + Cl2(s) Na+(aq) + e⁻ → Na(s)        E° = +0.362 V Cl2(s) + 2 e⁻  → 2 Cl-(aq)    E° = +0.789 V   use this equation: Δ G ° = − n F E cell °

Given the following information: F1- (aq): ΔGf° = -276.48 kJ/mol Na1+ (aq): ΔGf° = -261.87 kJ/mol 2 Na (s) + F2 (g) → 2 NaF (aq) a) Find ΔG° for this reaction using the above numbers. b) Write the expression for Qeq. (Note that NaF is a strong electrolyte!) c) Find ΔG at 325.0 K for 0.01500 atm F2 and 3.065 M NaF.

Consider the following reaction at 298K.2 Cu2+ (aq) + Co (s)  2 Cu+ (aq) + Co2+ (aq)Which of the following statements are correct?Choose all that apply. K < 1 n = 4 mol electrons The reaction is product-favored. Eocell > 0 delta Go > 0

For the following redox reaction calculate the Ecell and K at 298 K. 2Na+(aq) + Ca(s)→2Na(s) + Ca2+(aq) Half Reaction E(V) Na+(aq) + e- → Na(s) -2.71 Ca2+(aq) + 2e- → Ca(s) -2.87

Consider the following reductions. Cr2+(aq) + 2 e- ⟶⟶ Cr(s)                              Eored = - 0.910 V Co2+(aq) + 2 e- ⟶⟶ Co(s)                            Eored = - 0.280 V Which element is the better oxidizing agent? Explain your answer.

Calculate the K and AG for the following reaction at 25 °C: Zn(s) + CuSO₂(aq) → ZnSO₂(aq) + Cu (s) Note: Reference the Standard reduction potentials at 25 °C and Fundamental constants tables for additional information. Part 1 of 2 Round your answer to 3 significant digits. Bo = Part 2 of 2 AGⓇ ☐x10 = Round your answer to 4 significant digits. X kJ mol 0x10 3 X Ś

The free energy change for the following reaction at 25 °C, when [Cu2+] = 2.57x10-³ M and [I₂] = 1.10 M, is 88.8 kJ: 2Cu²+(2.57x10-3 M) + 2.001 (aq)—2Cu+ (aq) + 1₂(1.10 M) AG= 88.8 kJ What is the cell potential for the reaction as written under these conditions? Answer: V Would this reaction be spontaneous in the forward or the reverse direction?

Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: 2H+ (aq) + Hg(1)→ H₂(g) + Hg²+ (aq) 2H+ (aq) + 2e¯ → H₂(g) Fº red 2+ Hg2 (aq) + 2e¯ → Hg(1) AGO = || kJ K for this reaction would be Eº red = 0.000 V = = 0.855 V than one.