Given: Pb2+ (aq) + 2e- Pb(s); E degree = -0.13 V 2H+ (aq) + 2e- H2(g); E degree

Question

Given: Pb2+ (aq) + 2e- Pb(s); E degree = -0.13 V 2H+ (aq) + 2e- H2(g); E degree = 0.00 V NO3-(aq) + 4H+ (aq) + 3e- NO(g) + 2H2O(l); E degree = 0.96 V O2(g) + 4H+ (aq) + 4e- 2H2O(l); E degree = 1.23 V PbO2(s) + SO42-(aq) + 4H+ (aq) + 2e- PbSO4(s) + 2H2O(l); E degree = 1.69 V Under standard-state conditions, which of the following is the best oxidizing agent? O2 H+ NO3- Pb2+ PbO2 Which of the following statements is true concerning the electrochemical cell depicted below? Cu | Cu2+(aq) || Zn2+(aq) | Zn Cu2+(ag) + 2e- Cu(s); E degree = -2.38 V Zn2+(aq) + 2e- Zn(s); E degree = -2.87 V The cell reaction is spontaneous with a standard cell potential of 0.49 V. The cell reaction is nonspontaneous with a standard cell potential of -5.25 V. The cell is at equilibrium. The cell reaction is spontaneous with a standard cell potential of 5.25 V. The cell reaction is nonspontaneous with a standard cell potential of -0.49 V. Consider the following cell reaction: 2Ag+(aq) + H2(g) rightarrow 2H+(aq) + 2Ag(s); E degree cell = 0.80 V Under standard-state conditions, what is E degree for the following half-reaction? Ag+(aq) + 2e- rightarrow Ag(s) -0.80 V 0.40 V 0.80 V -0.40 V 1.10 V

Explanation / Answer

13.E.

pbO2 is reduced . The one gets reduced easily is the good oxidising agent

14. A

the reaction with more negative potential will be the anode. and the potential =Ecathode-E anode=0.49V

15.C.0.80V

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