1.
∆G=-nFE°
n =6, F=96485, E°=(-0.13_-1.66),=1.53V
∆G=-(6*96485*1.53)
=-886KJ
2.
n=2,F=96485, E°=(-0.4_-0.76)=0.36V
∆G=-(2*96485*0.36)
=-69.5KJ
Use the standard half-cell potentials listed below to calculate the standard free energy(K]for the following reaction...
Use the standard half-cell potentials listed below to calculate the standard cell potential and standard free energy Gº) for the following reaction occurring in an electrochemical cell at 25 °C. (The equation is balanced.) k(aq) + e-K(5) E* --2.93 V 12(s) + 2 0 - 2 (aq) E*-+0.54V a. +6.40 V & 670 KJ b. +1.85 V & 487K] C. 3.47 V & 241 Kb d. 3.47 V8 - 6709 e.+5,32 V &-+670K)
Use the half-reactions below to produce a voltaic cell with the given standard cell potential. Standard Cell Potential Co- (aq) + e-Cot (aq) E = +1.82 V 1.53 V 2H(aq) + 2e-H2(g) E = +0.00 V Pb2+ (aq) + 2e-Pb(s) E = -0.13 V Fe (aq) + e-Fel+ (aq) E = +0.77 V Ag (aq) + e-Ag(s) E = +0.80 V Sn* (aq) + 2e Sne (aq) 20.13 V Cu- (aq) + e- Cu(aq) E = +0.15 V Zn²+ (aq)...
Use the tabulated half-cell potentials below to calculate AG ( in Kilojoules for the following balanced redox reaction. F= 96,500 J/. mole e Pb2+(aq) + Cu(s) - Pb(s) + Cu2+(aq) Cu2+ (aq) + 2e -> Cu(s) E". Cu = -0.34 Volt Pb2+ (aq) + 2e-> Pb(s) E,Pb = -0.13 volt a. -41 kJ b. -81 kJ C. +46 kJ d. +91 kl e..91 kb 24 Calculate AS®rxn for the following reaction. The S' for each species is shown below the...
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...
Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: Pb2+(aq) + H2(g) Pb(s) + 2H+ (aq) Answer: kJ K for this reaction would be than one. Submit Answer Ret greater 4 more group attempts remaining less TU IOF LAS question. Use standard reductibn potentials to calculate the standard free energy change in kJ for the reaction: Pb2+ (aq) + Zn(s) Pb(s) + Zn?" (aq) Answer: K for this reaction would be than one....
14. A galvanic cell is composed of these two half cells, with the standard reduction potentials shown Co" (aq) + 2e = Co(s) -0.28 volt Cro(aq) + 3 e = Cr(s) -0.74 volt What is the standard free energy for the cell reaction of this galvanic cell? a. -88.8 kJ b. -178 kJ c. -266 kJ d. -295 kJ e. -590 kJ MO die W oss som sa robi bo bolo Videos onto 15. A galvanic cell is composed of...
Use the standard half-cell potentials listed below to determine which the following metals will dissolve in hydrochloric acid. Cl2(g) + 2e + 2C1-(aq); E° = 1.36 V 2H+(aq) + 2e + H2(g); E° = 0.00 V O Cu; E°(Cu2+/Cu) = +0.34V O Au; E°(Au3+/Au) = +1.50V O Al; E°(A13+/Al) = -1.66V O Ag; E°(Ag+/Ag) = +0.80V O Pt; E°(Pt2+/Pt) = +1.19V
Use the standard half-cell potentials listed below to determine which the following metals will dissolve in hydrochloric acid. Cl2(g) + 2e -- 2014(aq); E° = 1.36 V 2H+(aq) + 2e - H2(g); E° = 0.00 V OPt; E°(P+2+/Pt) = +1.19V O Ag; E°(Ag+/Ag) = +0.80V O Au; EⓇ(Au3+/Au) = +1.50V O Al; E(A13+/Al) = -1.66V O Cu; E(Cu2+/Cu) = +0.34V
19 20 Question 16 Half-cell Potentials: Half Reaction: E' value + 0.80 V +0.77 V Agte → AS Fe3+ + + Fe2+ Cu2+ 2e → Cu Pb2+ + 2e → Pb +0.34 V -0.13 V Ni2+ + 2e → NI -0.25 V - 0.40 V Cd2+ +2e → ca Fe2+ + 2e → Fe Zn2+ + 2e → Zn -0.44 V - 0.76 V A13+ +3 → AI - 1.66 V Consider an electrochemical cell constructed from the following half...
Use standard reduction potentials to calculate the standard free energy change (ΔGº) for the reaction below at 25º. Report your answer in kilojoules. 2 Cr+3 (aq) + 3 Co (s) ⟶ 2 Cr (s) + 3 Co+2 (aq)