A lithium battery is being developed. Its voltaic cell is constructed based on Li metal electrode and PbsO4 solid electrode.Their standard reduction potentials at 25 °C are given below.E°(Li* (aq) → Li (s)) = -3.040 VoltE°(PbSO4 (S) > Pb (s) + SO42 (aq) = - 0.356 Volt.Which of the following statements is true about this working (spontaneous) voltaic cell?a. The anode is PbsO4 (s) electrode.The standard cell potential of this voltaic cell is E°(cell) = 2.328 VoltO. PBSO4 (s) is a weaker oxidizing agent than Lit.To make a commercial product, the voltaic cell is adjusted to have [Lit] = 2.1 M and [SO42] = 3.5 M. The celld.potential E under this condition should be less than the standard cell potential E°(cell).e. None of the above

We are given standard reduction potential values of two different half cells:

A lithium battery is being developed. Its voltaic cell is constructed based on Li metal electrode and PbSO4 solid electrode. Their standard reduction potentials at 25 °C are given below. E°(Li* (aq) > Li (s)) =-3.040 Volt E°(PBSO4 (s) → Pb (s) + So42 (aq)) = - 0.356 Volt. Which of the following statements is true about this working (spontaneous) voltaic cell? O a. The anode is PbSO4 (s) electrode. The standard cell potential of this voltaic cell is E°(cell) = 2.328 Volt c. PBSO4 (s) is a weaker oxidizing agent than Lit. To make a commercial product, the voltaic cell is adjusted to have [Lit] = 2.1 M and [SO42] = 3.5 M. The cell potential E under this condition should be less than the standard cell potential E°(cell). e. None of the above

A lithium battery is being developed. Its voltaic cell is constructed based on Li metal electrode and PbSO4 solid electrode. Their standard reduction potentials at 25 °C are given below. E°(Li* (aq) > Li (s)) =-3.040 Volt E°(PBSO4 (s) → Pb (s) + So42 (aq)) = - 0.356 Volt. Which of the following statements is true about this working (spontaneous) voltaic cell? O a. The anode is PbSO4 (s) electrode. The standard cell potential of this voltaic cell is E°(cell) = 2.328 Volt c. PBSO4 (s) is a weaker oxidizing agent than Lit. To make a commercial product, the voltaic cell is adjusted to have [Lit] = 2.1 M and [SO42] = 3.5 M. The cell potential E under this condition should be less than the standard cell potential E°(cell). e. None of the above

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

Please don't provide handwritten solution ...
The potential of the following voltaic cell is 0.55 V. Given this information, determinethe Kb for the organic base (RNH2) dissolved in the left cell.
Give only typing answer with explanation and conclusion
Consider the following voltaic cellat 25°C: Al(s) | Al3*(0.20 M, aq) || Al3+(0.75 M, aq) | Al(s) If the standard reduction potential for the Al3+(aq) /Al(s) redox couple is -1.66 V, what is the overall cell reaction and the cell potential of the voltaic cell shown above? (Reference the cell as written above.) The overall cell reaction is Al3+(0.75 M, aq) AIS*(0.20 M, ag) and E°cell = +0.0113 V. There is no overall cell reaction and E°cell = 0.00 V. The overall cell reaction is Al3+(0.20 M, aq) Als*(0.75 M, aq) and E°cell = +0.0113 V. The overall cell reaction is Al3+(0.20 M, aq) →A13+(0.75 M, ag) and E°cell| = -0.0113 V. The overall cell reaction is Al3*(0.75 M, aq) →AI3*(0.20 M, aq) and E°cell = -0.0113 V. None of these
(ii) State which species is the oxidising agent.
-3 M solution of Cu(NO3)2. The other half-cell consists of a zinc electrode in One half-cell in a voltaic cell is constructed from a copper wire electrode in a 3.9 x 10 a 0.45 M solution of Zn(NO3)2 . Calculate the cell potential. You may need to use the following data: Zn²+ (aq) + 2 e¯ → Zn(s) E° = -0.763 V Cu²+ (aq) + 2 e¯ → Cu(s) E° = +0.337 V Potential = V
A chemist designs a galvanic cell that uses these two half-reactions: half-reaction standard reduction potential 2+ Zn (aq)+2e - Zn(s) :-0.763 V (red |Cro (aq)+4 H2O(1)+3e Cr(OH)3(s)+5OH (aq) :-0.13 V = - (red Answer the following questions about this cell. Write a balanced equation for the half-reaction that happens at the cathode. x10 Write a balanced equation for the half-reaction that happens at the anode. Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous as written.
5. A more complex example where other atoms are involved. Balance the following reaction in basic solution. r (aq) + MnO4 (aq) MnO, (s)+ BrO;" (aq) TABLE 19.1 Standard Electrode Potentials at 25 °C Reduction Half-Reaction E'(V) F + 20 H,0,(aq) + 2 H(aq)+2 e 2F(aq) 2.87 Stronger oxidizing agent Weaker 2 H,0 - PBSO (s) +2 H,0) -Mn0(s)+ 2 H,0) Mn (aq) 4 H,O) - Auts) 1.78 reducing agent PO,(s) + 4 H"(ag) + S0,2-(aq) + 2 e 1.69 Mno, (aq)+ 4 H'(aq) 3 e Mno, (aa) + 8 H (aq)+5 e Au (aq)+3e PbO,(s)+ 4 H(aa) + 2 e Cl,(g) + 2 e" 1.68 1.51 1.50 1.46 - P (ag) + 2 HO - 2a(aq) 1.36 1.33 Cr0, (aq) + 14 H'(ag) + 6 e 0+ 4 H (aq) + 4 e MnO(s)+4 H'(aq) +2 e 10, (aq) 6 H'(aq) + 5e Br+2e vo, (aq) + 2 H'(aq) + e NO, (aq) + 4 H°(aq) + 3e CIO+e 2 C"(aq) + 7 H,On - 2 H,O1) 1.23 Mn (aq) + 2 HyOV) 121 ulae) + 3 H,0() 2 Br (ag) - vo"(aq) +H,00 - NO + 2 H,O0 - CIO, (aa) - Agis) 1.20 1.09 1.00 0.96 0.95 Ag"(aa) + e 0.80 Fe (aq) + O+2 H'(aq)+2e - Fe*(aq) 0.77 - H,O,(ag) 0.70 Mn0, Taa) + e Mn0 (aa) 0.56 8) + 2 e…
(c) A galvanic cell is constructed using the following two half-cells at 25°C and 1 atm. The concentration of each species is given in the half-cell equation. Fe* (aq, 1.10 mol dm³) +3e → Fe2+ (aq, 0.86 mol dm³) E° = +0.77 V Ag* (aq, 1.35 mol dm³) + e → Ag(s) E° = +0.88 V
here is the full question. can you please please not reject it and answer
Consider the following reduction potentials. Ag*(aq) + e → Ag(s) E° = + 0.80 V Sn2*(aq) + 2 e -→Sn(s) E° = - 0.14 V Mg2 (aq) + 2 e - → Mg(s) E° = - 2.37 V a.) Calculate the cell potential at standard conditions (E°cell) for a voltaic cell of Mg2+/Mg and Ag*/Ag, in Volts. E° cell = v b.) Calculate Ecel in V at 298 K if [Mg2*] = 0.100 M and [Ag*] = 0.100 M. %3D Ecell = V
Given the following two half-reactions, write the overall reaction in the direction in which it is product- favored, and calculate the standard cell potential. Pb2*(aq) + 2 e- - → Pb(s) E° = -0.126 V Fe3*(aq) + e¯ → Fe2*(s) E° = +0.771 V Pb2*(aq) + 2 Fe²*(s) → Pb(s) + 2 Fe³*(aq); = +0.897 V cell Pb2*(aq) + Fe²*(s) → Pb(s) + Fe³*(aq); E, cell = +0.645 V Pb(s) + 2 Fe³*(aq) → Pb²*(aq) + 2 Fe²*(s); F° cell = +1.416 V Pb(s) + 2 Fe*(aq) → Pb²*(aq) + 2 Fe²*(s); E. 'cell = +0.897 V Pb(s) + Fe³*(aq) → Pb²*(aq) + Fe²*(s); = +0.645 V