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Question 56 (1 point) Use the standard half-cell potentials listed below to calculate the standard cell...
Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring in an electrochemical 3 C12(a) + 2 Fe(s) - 6 Cr(aq) + 2 Fe3+ (aq) Cl2(g) + 2 + 2 Cl(aq) Fe3+ (aq) + 36 - Fe(s) E' = +1.36 V E' = -0.04 V +1.40 V O-1.40 V 0 +4.16 V O +1.32 V O-1.32 V Submit Resvest Answer
24) Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 25°C. (The equation is balanced.) 26). 3 Cl2(8) +2 Fe(s) - 6 Cl(aq) + 2 Fe3+ (aq) 12(8)+2 e 2Cl(aq) E° 1.36 V 3C1tbe ! Fe3+(aq) + 3 e → Fe(s) 80036V 2Fe > 25€ Ep 3 + the A) 240 V B) 4.16V ) -1.32 V D) -1.40 V E) 1.32 V
Question 52 (1 point) For a reaction Keq - 1.2 x 10-6 at T = 298 K. What is AG for the reaction? -53.6 kJ/mol 53.6 kJ/mol 0-22.7 kJ/mol 33.8 kJ/mol 22.7 kJ/mol Question 47 (1 point) Calculate the pH of a solution that is 0.310 Min sodium formate (NaHCO2) and 0.190M in formic acid (HCO2H). The Kof formic acid is 1.77 x 10-4. 13.79 3.532 3.958 O 4.975 10.04
Question 55 (1 point) Calculate AGrxn at 398 K under the nonstandard conditions shown below for the following reaction. SO3(8) + H2O(g) + H2SO4(1) AG°- -90.5 kJ P(SO3) = 0.20 atm, P(H2O) = 0.88 atm 0-93.2 kJ +93.2 kJ -86.2 kJ -84.8kJ +87.8 kJ
Question 47 (1 point) Calculate the pH of a solution that is 0.310 Min sodium formate (NaHCO2) and 0.190M in formic acid (HCO2H). The Ką of formic acid is 1.77 x 10-4. 13.79 3.532 3.958 4.975 10.04 MOCRA Question 42 (1 point) What is the hydronium ion concentration of a 0.100 M hypochlorous acid solution with Ka = 35 x 10-87 The equation for the dissociation of hypochlorous acid is: HOCH(aq) + H2011) =H30*(aq) + OCH@g). O 1.9 x 10-5...
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 standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 25�C. (The equation is balanced.) Sn(s) + 2 Ag+(aq) ? Sn2+(aq) + 2 Ag(s) Sn2+(aq) + 2 e- ? Sn(s) E� = -0.14 V Ag+(aq) + e- ? Ag(s) E� = +0.80 V a)-1.08 V b)+1.74 V c)-1.74 V d)+0.94 V e)+1.08 V
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
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)...