EMF, Nernst equation and applications
- Find the emf of the cell in which the following reaction takes place at 298 KNi(s) + 2Ag^+ (0.001\ M) → Ni^2+ (0.001\ M) + 2Ag(s)(Given that E^°_cell = 10.5 V, 2.303 RT/F = 0.059 at 298 K) Answer: Dropped in the official answer key: full marks to every candidate Key idea: use the Nernst equation E=E^°-0.059/nlog Q . Reaction: m... 1
- Given below are half cell reactions :MnO_4^- + 8H^+ + 5e^- → Mn^2+ + 4H_2O,E^°_Mn^2+/MnO_4^- = -1.510 V1/2O_2 + 2H^+ + 2e^- → H_2O,E^°_O_2/H_2O = +1.223 VWill the permanganate ion, MnO_4^- liberate O_2 from water in the presence of an acid ? Answer: 2. Yes, because E^°_cell = +0.287 V Key idea: a reaction is spontaneous if E^°_cell>0 , using the reduction potentials of the two ha... 1
- Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below :BrO_4^- 1.82\ V BrO_3^- 1.5\ V HBrO 1.595\ V Br_2 1.0652\ V Br^-Then the species undergoing disproportionation is Answer: 4. HBrO A species disproportionates if the emf for its reduction (to the species on one side) is higher than the emf for its oxidati... 1
- In the electrochemical cell :Zn|ZnSO_4(0.01\ M)||CuSO_4(1.0\ M)|Cu, the emf of this Daniel cell is E_1. When the concentration of ZnSO_4 is changed to 1.0 M and that of CuSO_4 changed to 0.01 M, the emf changes to E_2. From the followings, which one… Answer: 2. E₁ > E₂ Key idea: apply the Nernst equation, E = E^° - 0.059/2logdfrac[Zn²⁺][Cu²⁺] . E₁ : log0.01/1=-2 , so E₁ = E^° + 0.059 . E₂... 1
- The pressure of H_2 required to make the potential of H_2 - electrode zero in pure water at 298 K is : Answer: 1. 10⁻¹⁴ atm Key idea: for the half cell 2H^++2e^-→ H₂ , the Nernst equation is E=0-0.059/2logfracP_(H_2)[H^+]² In pure water [H^+]=... 1