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14. Calculate Ecal and the equilibrium constant, K, for the following reaction at 25°C. Cd(s) +...
6) Calculate the equilibrium constant K at 25°C for the the equilibrium constant K at 25°C for the following reaction for the standard cell potential: (points) (AFE-R7 In K, F-96485 Címol,R-8.37 J/molk) Pb2+ (aq) + Fe(s) S Pb(s) + Fe?*(aq) 7) Calculate the cell potential of the following cell at 25°C. (7 points) Fe(s)| Fe?"(aq) (1.1 M) || Cu?"(aq) (0.50 M) Cu(3) Ecall-E Call - 0.0592/n logo
6) Calculate the equilibrium constant K at 25°C for the following reaction for tant K at 25°C for the following reaction for the standard cell potential (7 points) (nFEⓇ - RT In K, F=96485 C/mol.R=8.31 J/molk) Pb2+ (aq) + Fe(s) 5 Pb(8) + Fe²(aq) 7) Calculate the cell potential of the following cell at 25°C. (7 points) Fe(s) | Fe*(aq) (1.1 M) || Cu?" (aq) (0.50 M) Cu() Ecell - Eºcell = 0.0592/n logQ
Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Fe+ (aq) + Cd(s)— 2Fe2+ (aq) + Cd²+(aq) Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. Equilibrium constant AGⓇ for this reaction would be than zero. Submit Answer Retry Entire Group 6 more group attempts remaining Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Cu + (aq) - Cu(s)— 2Cu (aq) + Cu+...
cell = 0.74 V. Calculate the equilibrium constant, K. For the following redox reaction at 25 °C, E° Cd(s) + Cu 2+ (aq) → Cd 2+ (aq) + Cu(s)
The equilibrium constant, K, for a redox reaction is related to the standard potential, Eº, by the equation In K = nFE° RT where n is the number of moles of electrons transferred, F (the Faraday constant) is equal to 96,500 C/(mol e), R (the gas constant) is equal to 8.314 J/(mol · K), and T is the Kelvin temperature. Standard reduction potentials Reduction half-reaction E° (V) Ag+ (aq) + e +Ag(s) 0.80 Cu²+ (aq) + 2e + Cu(s) 0.34...
1. Which metal can reduce Ni2+ ions but not Fe2+ ions?
Cd²⁺,Cr,Sn²⁺,Sn,Cd
2.Which species can oxidize Sn but not Fe?
Ni,Ni²⁺,Pb²⁺,Cd,Pb
1.68 1.51 1.50 1.09 1.00 0.96 0.95 FAO) + 2 H,0,1a) + 2H(aq) + 2e PO4 + 4H"(aq) + S0,00) + 2€ Mno. (aq) + 4H(aq) + 3e Mno,aq) + 8H(aq) + 5e Au?+ (aq) + 3€ PbO (8) + 4H(aq) + 2 Cl@) + 2 Cr,0,?" (aq) + 14 H*(aq) + 6e 0,(@) + 4H(aq) + 4e MnO...
A) Use tabulated electrode potentials to calculate ΔG∘ for the reaction. 2K(s)+2H2O(l)→H2(g)+2OH−(aq)+2K+(aq) B) (Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V) An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:Ni(s)VO2+(aq,0.024M)+2H+(aq,1.4M)+e−→→Ni2+(aq,1.8M)+2e−VO2+(aq,1.8M)+H2O(l) Calculate the cell potential under these nonstandard concentrations. C) Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V ) Half-reaction E∘ (V ) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) − 0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) − 0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) − 0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) − 0.76...
se the tabulated half-cell potentials to calculate ??-for the following ,edo-m-mm in (7 points ) Tabulated half-cell potentials (Reduction) Zar" (aq) + 2 e-? Zn(s) E.--0763 V Cl:(g) +2e2 CT(a)+1.358 F-96,485 C/mol. electron, 13-1 C.V and 1kJ -1000 C). (7 points) 4) Calculate K for the oxidation of copper by H' (at 25 Cu(s) + 2 r(aq) ? Cu2+(aq) + H2(g) Tabulated half-cell potentials (Reduction) Cu2 (aq) + 2 e-Al(s) E-0.34 v 2H+(aq) + 2 e--+ H2 (g) E#20.00 V...
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6. Calculate the equilibrium constant at 25°C for the following reaction: Cd(s) + Sn** (aq) → Cd²+ (aq) + Sn(s) K = 6.3 x 108
please help, thanks!
7. What is the correct cell notation for the reaction below? Cd(aq)+Ni(s)-Cd(s)+ Ni(aq) a. Cd Cd| Ni | Ni d. Ni Cd I N | Ca b. Cd CdNi | Ni e. Cd |Cd Ni Ni 8. Calculate AS for the dissociation of dinitrogen tetraoxi de at 25°C NOdg) 2NO:(g) AG, J/mol)AH, J/mol) Substance NO) NOdg) +33.2 +51.3 +97.9 +9.2 d. +0.076 kJ K a.-2.10 kJ/K e +2.10kJ/K b. -0.550 kJ/K c.-0.208 kJ/K 9. Use the standard...