10. Calculate the Ksp of lead(II) iodide from the following standard electrode roper electrode potentials from...
Calculate the equilibrium constant for each of the reactions at 25 ∘C. Standard Electrode Potentials at 25 ∘C Reduction Half-Reaction E∘(V) Pb2+(aq)+2e− →Pb(s) -0.13 Zn2+(aq)+2e− →Zn(s) -0.76 Br2(l)+2e− →2Br−(aq) 1.09 Cl2(g)+2e− →2Cl−(aq) 1.36 MnO2(s)+4H+(aq)+2e− →Mn2+(aq)+2H2O(l) 1.21 Pb2+(aq)+2e− →Pb(s) -0.13 Br2(l)+2Cl−(aq)→2Br−(aq)+Cl2(g) Express your answer using two significant figures.
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...
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)...
Calculate the equilibrium constant for each of the reactions at 25∘C∘C. Standard Electrode Potentials at 25 ∘C Reduction Half-Reaction Cl2(g)+2e ---> 2CI- 1.36 I2(s)+2e --> 2I- 0.54 Part A Cl2(g)+2I−(aq)→2Cl−(aq)+I2(s) K= ?
4. (a) Use the standard reduction potentials at 25° C in Table 18.1 in Tro, Fridgen and Shaw, and calculate the standard emf E° of an electrochemical cell described by the following reaction: 3 Zn + 2 Cr3+ + 2 Cr + 3 Zn? (b) What is n? (c) What is AGº for this reaction at 25°? (d) What is the equilibrium constant for this reaction at 25°? TABLE 18.1 Standard Reduction Potentials at 25°C EV) 2.87 1.61 1.51 1.36...
What is the standard emf of a galvanic cell made of a Co electrode in a 1.0 M Co(NO32 solution and a Al electrode in a 1.0 M AI(NO3)3 solution at 25°C? 0 cell Standard Reduction Potentials at 25°C Half-Reaction E(V +2.87 +2.07 +1.82 O,(g) 2H (aq)2e0(g)+HO Co3+(aq) + e-_? Co2+(aq) H,02(aq) + 2H"(aq) + 2e-_ 2H20 Cu2+(aq) + 2e-? Cu(s) AgCIs) + Ag(s) + CI(a) S02-(aq) + 4H'(aq) + 2e S02(g) + 2H20 Cu2+(aq) + e-_ Cu+(aq) Sn (aq)...
Based on the information in the table of standard reduction potentials below, what is the standard cell potential for a galvanic cell that has sodium and copper electrodes immersed in 1M Nat and Cu2+ solutions? Also, identify the cathode. Half-reaction E° (V) Aut te - Au +1.69 I2 + 2e - + 21 +0.54 Cu²+ + 2e - Cu +0.34 → Fe -0.04 Feit +3e Zn2+ + 2e -0.76 - Zn -2.71 Natte Na 0 -3.05 V Na is the...
Standard reduction half-cell potentials at 25°C E (V) E (V) 1.50 -0.45 0.80 -0.50 0.77 -0.73 0.52 -0.76 0.34 -1.18 Half-reaction Aut (aq) + 3e +Au(s) Ag+ (aq) + +Ag(s) Fe3+ (aq) +34 Fo+ (aq) Cut(aq) + Cu(s) Cu²+ (aq) + 2e +Cu(s) 2H+ (aq) - 2e +H2 (6) Fe3+ (aq) + 3e Fe(s) Pb2+ (aq) + 2e →Pb(s) Sn-(aq) + 2e +Sn(s) Ni2+ (aq) + 2e →Ni(s) Co2(aq) +2e + Co(s) ca? (aq) + 2e +Cd(s) 0.00 Half-reaction Fe(aq)...
Calculate the equilibrium constant for each of the reactions at 25∘C. Standard Electrode Potentials at 25 ∘C Reduction Half-Reaction E∘(V) Fe3+(aq)+3e− →Fe(s) -0.036 Sn2+(aq)+2e− →Sn(s) -0.14 Ni2+(aq)+2e− →Ni(s) -0.23 O2(g)+2H2O(l)+4e− →4OH−(aq) 0.40 Br2(l)+2e− →2Br− 1.09 I2(s)+2e− →2I− 0.54 A) 2Fe3+(aq)+3Sn(s)→2Fe(s)+3Sn2+(aq) (answers are not 4.1x10^5, 3.3x10^3, 2.7x10^10, or 2.6x10^10) B) O2(g)+2H2O(l)+2Ni(s)→4OH−(aq)+2Ni2+(aq) C) Br2(l)+2I−(aq)→2Br−(aq)+I2(s) (answer is not 1.7x10^18)
Write the half reactions and overall reaction for each cell with calculated overall potentials as shown in Table 5-1. (Note: for the iron solutions the Nernst equation must be used) Pb(s) | Pb(NO3)2 (0.1M) || Cu(NO3)2 (0.1M) Cu(s) Zn(s) | Zn(NO3)2(0.1M) || Cu(NO3)2 (0.1M) Cu(s) Cds) | Ca(NO3)2 (0.1M) || Cu(NO3)2 (0.1M) | Cu(s) Cu() Cu(NO3)2(0.1M) Il Fe (0.1M/Fe? (0.1M graphite Pb(s) Pb(NO3)2(0.1M) Il Fe3(aq) (0.1M)/ Fe2(aq) (0.1MI graphite(s) Zns | Zn(NO3)2 (0.1M) || Pb(NO3)2 (0.1M) | Pb(s) Cdis Ca(NO3)2...