The following reaction is not associated with NiMH batteries A. MH + OH- ⇋ M + H2O + e- B. Ni(OH)2 + OH- ⇋ NiOOH + H2O + e- C. 2H2O + 2e- H2+ 2OH- D. 2H+ + 2e ⇋ H2
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The following reaction is not associated with NiMH batteries
A. MH + OH- ⇋ M + H2O + e-
B. Ni(OH)2 + OH- ⇋ NiOOH + H2O + e-
C. 2H2O + 2e- H2+ 2OH-
D. 2H+ + 2e ⇋ H2
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- Calculate E0 that would be developed by the cells using the following reactions (all concentrations 1M) 1. Ni + 2Ag+↔ Ni2+ + 2Ag2. 2Cr + 3Cu2+ ↔ 2Cr3+ + 3CuThe E°cell = 0.135 V for the reaction3I2(s) + 5Cr2O72-(aq) + 34H+(aq) → 6IO3-(aq) + 10Cr3+(aq) + 17H2OWhat is Ecell if [Cr2O72-] = 0.014 M, [H+] = 0.21 M, [IO3-] = 0.00018 M, and [Cr3+] = 0.0039 M?Ecell =From the below data: Please show a detail on how to calculate Q and Ecell (calc) Ecell = 0.941V Zn2+(aq)+2e- → Zn(s) E° = -0.76 Cu2+(aq)+2e- → Cu(s). E° = +0.34 E°cell = 0.76+0.34 = 1.1V Balanced chemical reaction Zn(s) + Cu²+(aq) --> Cu(s) + Zn²+(aq)
- Calculate emf of the following cell at 25 °C:Fe | Fe2+(0.001 M) 11 H+(0.01 M) | H2(g) (1 bar) | Pt(s)E°(Fe2+|Fe) = -0.44V, E°(H+|H2) = 0.00 V1) Calculate Delta G° for the overall reaction at 25.0 °C 2) Calculate the E° for the anode at 25.0 °C.acid dissociation constants for H3PO4 (Ka1 = 7.11 × 10-3, Ka2 = 6.32 × 10-8, Ka3 = 7.1× 10-14), calculate standard reduction potential for the half reactionH2PO4- + H+ + 2e HPO32- + H2O
- Below are the reduction half reactions for chemolithoautotrophic nitrification, where ammonia is a source of electrons and energy and oxygen is the terminal electron acceptor. NO2- + 6e- -> NH4+ (E0 = +0.34 volts) O2 + 4e- -> 2H2O (E0 = +0.82 volts) If you balance and combine the reactions so that 293 moles of NH4+are oxidized to NO2-, how many moles of electrons will be transferred from Nitrogen to Oxygen?What reactions occur during the electrolysis of aqueous Nal between Pt electrodes? ε° of Na | Na+ = -2.71 V ε° of Pt | I2, I- = 0.53 V ε° of Pt | H2O, H2, OH- = -0.83 V ε° of Pt | H2O, O2, H+ = 1.23 V a. Na (s) deposits at the cathode, O2 (g) evolves at the anode b. Na (s) deposits at the cathode, I2 (s) deposits at the anode c. H2 (g) evolves at the cathode, I2 (s) deposits at the anode d. H2 (g) evolves at the cathode, O2 (g) evolves at the anodeIf Ecell=-0.36V for the reaction below at 25C, what is the value of the equilibrium constant at this temperture. Zn2+(aq) + Cd(s) = Cd2+(aq) + Zn(s).
- Calculate the cell potential of the indicated reaction as shown below. Ni(OH)2 <-------> Ni2+ + 2OH- Ksp = 1.6 x 10-16 Ni2+ + 2e- --------> Ni Eo = -0.23 V Ni + 2OH- ---------> Ni(OH)2 + 2e- Eo = ? Use:F = 96485 Coulombs/mole ; R = 8.31446 Answer must be in 3 decimal places.What reactions occur during the electrolysis of aqueous Nal between Pt electrodes? ε° of Na | Na+ = -2.71 V ε° of Pt | I2, I- = 0.53 V ε° of Pt | H2O, H2, OH- = -0.83 V ε° of Pt | H2O, O2, H+ = 1.23 VGiven a cell: 2 Ag+(aq) + Cu(s) ----> 2Ag (s) + Cu2+ (aq), E(cell) = 0.346 V, E0(cell) = E0cathode - E0anode = 0.800 V - 0.340 V = 0.460 V, I. calculate the concentration of Cu2+ given that of Ag+ is 1.0 x 10-5M Cu2+M= Keq =