13. Ca2+(aq)+2e - Ca(s)E.=-2.84V 2Li(s) - 2Li+(aq)+2e-E-=-3.040V Where is the location of oxidation and where will b...
Calculate the cell potential, the equilibrium constant, and the free-energy change for: Ca(s)+Mn2+(aq)(1M)⇌Ca2+(aq)(1M)+Mn(s) given the following Eo values: Ca2+(aq)+2e−→Ca(s) Eo = -1.59 V Mn2+(aq)+2e−→Mn(s) Eo = -0.54 V 1.) Calculate the equilibrium constant. 2.) Free-energy change?
Calculate the value of Eºcell for the reaction 2Ag(s) + Ca2+(aq) + 2Ag+(aq) + Ca(s), using the following standard reduction potentials. Ag+(aq) + e- → Ag(s) E° = 0.80 V Ca2+(aq) + 2e + Ca(s) E° = -2.87 V 2.07 V -3.67 V 3.67 V -2.29 V -2.07 V
Given the following reduction half-reactions: Fe3+(aq)+e−→Fe2+(aq) E∘red=+0.77V S2O2−6(aq)+4H+(aq)+2e−→2H2SO3(aq) E∘red=+0.60V N2O(g)+2H+(aq)+2e−→N2(g)+H2O(l) E∘red=−1.77V VO+2(aq)+2H+(aq)+e−→VO2+(aq)+H2O(l) E∘red=+1.00V Write balanced chemical equation for the oxidation of Fe2+(aq) by S2O2−6(aq). Calculate ΔG∘ for this reaction at 298 K. Calculate the equilibrium constant Kfor this reaction at 298 K. Write balanced chemical equation for the oxidation of Fe2+(aq) by N2O(g). Calculate ΔG∘ for this reaction at 298 K. Calculate the equilibrium constant Kfor this reaction at 298 K. Write balanced chemical equation for the oxidation of Fe2+(aq)...
Consider the following cell reaction at 18°C: Ca(e)+Cu+ (aq) + Ca2+ (aq) + Cu() Calculate the standard cell potential of this cell from the standard electrode potentials, and from this, obtain AG" for the cell reaction. Calculate AF. Use these values of AN and AG to obtain AS for the cell reaction. Ca²+ (aq) +20 + Ca() --2.76 V Cu? (g) +20 + Cu(s) - 0.84 V AH;(O.*()) -- 542.8 kJ/mol AH;(Out (as)) - 64.8 kJ/mol V AG- AH- kJ...
Calculate the cell potential for Ca(s) + Mn2+(aq) ↔ Ca2+(aq) + Mn(s). Assume any aqueous species has a concentration of 1 M. Ca2+(aq) + 2e- → Ca(s) Eº = -2.87 V Mn2+(aq) + 2e- → Mn(s) Eº = -1.18 V
Calculate the value of K given the following information anodel:(oxidation) Fe(s) Fe2+ (aq) + 2e E ? Fe = -0.447 V cathodel:(reduction): 2 x (Ag+ (aq) + e- Ag(s) Eig'lag = 0.7996; V 0.0592 Hint: Use E cell logk Calculate E Cell first. n 2.6 O 1.247 O 1.3X1042 none of the above
Use the tabulated electrode potentials to calculate K for the oxidation of nickel by H+: Ni(s)+2H+(aq)→Ni2+(aq)+H2(g) Express your answer using two significant figures. 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 Cu2+(aq)+2e−→Cu(s) 0.34 Mn2+(aq)+2e−→Mn(s) −1.18 2H+(aq)+2e−→H2(g) 0.00 Al3+(aq)+3e−→Al(s) −1.66 Fe3+(aq)+3e−→Fe(s) −0.036 Mg2+(aq)+2e−→Mg(s) −2.37 Pb2+(aq)+2e−→Pb(s) −0.13 Na+(aq)+e−→Na(s) −2.71 Sn2+(aq)+2e−→Sn(s) −0.14 Ca2+(aq)+2e−→Ca(s) −2.76 Ni2+(aq)+2e−→Ni(s) −0.23 Ba2+(aq)+2e−→Ba(s) −2.90 Co2+(aq)+2e−→Co(s) −0.28 K+(aq)+e−→K(s) −2.92 Cd2+(aq)+2e−→Cd(s)...
of Write the balanced oxidation half-reaction for the following overall reaction: 2 H+ (aq) + Ca(s) → Ca2+(aq) + H2(g) Select one: a. H2(g) → 2 H (aq) + 2e Ob. 2 H(aq) - Halg) +2e c. none of the above d. Ca(s) + 2e + Calt (aq) MacBook Air
Calculate the cell potential, the equilibrium constant, and the free energy change for Ca(s) + Mn2+ (aq)(1M)=Ca²+ (aq)(1M) + Mn(s) given the following E' values: Ca2+ (aq) + 2e →Ca(8) Eº = -1.50 V Mn2+ (aq) + 2e →Mn(s) E° = -0.58 V
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...