sborie Il.A galvanic cell is based on the reaction priwbnbon MnO, (aq) + H*(aq) + CI0,-(aq) → CIO, (aq) + Mn2*(aq) + H2O(I) The standard half-reactions are anoins bos MnO4 + 5e + 8H+ Mn2+ + 4H,O = 1.51 V CIO, + 2H+ + 2e- → CIO3 + H2O Eo = 1.19 V a. What is being oxidized and what is being reduced? b. What is the oxidation reaction and What is the reduction reaction? C. What is the half reaction that occur at the cathode? d. Calculate the standard cell potential produced by the galvanic cell bssd ed 6 oibs s ot nejzil. 99loo Jorl gnilninb.a ert to Inemevom erb yd b9elgmos 2 o rbisw leigib ert

sborie Il.A galvanic cell is based on the reaction priwbnbon MnO, (aq) + H(aq) + CI0,-(aq) → CIO, (aq) + Mn2(aq) + H2O(I) The standard half-reactions are anoins bos MnO4 + 5e + 8H+ Mn2+ + 4H,O = 1.51 V CIO, + 2H+ + 2e- → CIO3 + H2O Eo = 1.19 V a. What is being oxidized and what is being reduced? b. What is the oxidation reaction and What is the reduction reaction? C. What is the half reaction that occur at the cathode? d. Calculate the standard cell potential produced by the galvanic cell bssd ed 6 oibs s ot nejzil. 99loo Jorl gnilninb.a ert to Inemevom erb yd b9elgmos 2 o rbisw leigib ert

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...

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MnO4– + Sn ----> MnO2 + Sn2+ A) Identify the half-reaction that occurs at the cathode. B) Identify the oxidizing agent and anode. C) standard cell potential? D) What will be the potential of a cell constructed using 57.0 g MnO2, 1.57 M MnO4–, 14.8 g Pt, 12.5 g Sn and 0.87 M Sn2+ at pH 2.5?
Standard Electrode Potentials at 25 °C Reduction Half-Reaction O Cu (aq) O H,Ozlaq) O Br O 10₂ (aq) Fyl) +20 H₂O₂(aq) + 2 H(aq) +2e" PbOg(s) +4 H(aq) +50, (aq) + 2 e Mn0 (aq) +4 H(aq) + 3 e MnO, (aq) +8H(aq) + 5 e Au³(aq) + 3 e PbOy(s) +4 H(aq) + 2e¯ C+20 00, (aq) +14 H(aq) + 6 e 00+ 4H*(aq) + 4 MnO₂ (s) +4 H(aq) +2e¯ 10₂(aq) + 6H(aq) +5e Big(0+20 VO₂ (aq) + 2 H(aq) + e¯ NO₂ (aq) + 4H(aq) + 3 e CO₂(0)+ Ag (ag)+e Fe(aq) + e 0)+2 H(aq) +2 e Mn0, (aq) + e (8) +26 Cu (aq) + e -2 (aq) -2 H₂O(0) Which of the following is the strongest oxidizing agent? -PbSO4(s) + 2 H₂O(0) →→MnO₂ (s) + 2 H₂O(0) Mn²(aq) + 4H₂O(1) - Au(s) ·Pb²+ (aq) + 2 H₂O(0) - 2 Cr (aq) 2 Cr³+ (aq) + 7 H₂O(1) → 2 H₂O( -2(aq) -Cu(s) Mn²+ (aq) + 2 H₂O(0) ₂(aq) + 3 H₂0(0) -2 Br (aq) - Vo²+ (aq) + H₂O( -NO(g) + 2 H₂O(1) — CIO; (aq) -Ag(s) -Fe²+ (aq) -) HyOz(ag) MnO₂ (aq) 1.00 0.96 0.95 0.80 0.77 0.70 0.56 0.54 0.52 E° (V) 2.87 1.78 1.69 1.68 1.51 1.50 1.46 1.36 1.33 1.23 1.21 1.20 1.09
(ii) State which species is the oxidising agent.
Consider the following half-reactions and their standard reduction potentials then give the standard line (cell) notation for a voltaic cell built on these half reactions. Mn2*(aq) + 2 e Mn(s) E° = -1.18 V Fe(aq) +3 e Fe(s) E° = -0.036 V A. Fe(s) | Fe3t(aq, 1.0 M) || Mn (s) | Mn 2*(aq. 1.0 M) B. Mn (s) | Mn 2*(aq, 1.0 M) | | Fe(s) | Fe(aq. 1.0 M) C. Fe(s) | Fe3t(aq. 1.0 M) || Mn2*(aq, 1.0 M) | Mn(s) D. Mn (s) | Mn 2*(aq. 1.0 M) || Fe(aq, 1.0 M) | Fe(s)
Calculate Eu for the following cell: Pt(s) | H₂(g) | H *(aq) || Pb2+ (aq) | PbSO4(s) | Pb(s) given the following standard reduction potentials. 2 H(aq) + 2e →→ H₂(g) PbSO4(s) + 2e →→ Pb(s) + SO42 (aq) O a. -0.712 V O b. +0.356 V O c. -0.356 V O d. +0.712 V O e. -0.178 V E° = 0.000 V E = -0.356 V
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…
Using the standard reduction potentials here, calculate the standard potential for the following cell in V (make sure you don't round your answer): Culs)Cu2"(aq)||Mn2 (aq).H"(aq)|MnO2(s)Pt(s) Your Answer:
(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
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
A voltaic cell is constructed with two silver electrodes, where the half-reaction isAg2+(aq) + 2e- → Ag (s) E° = -0.322 VDetermine the cell emf at 25 °C when the concentrations of chloride ion in the two compartments are 4.0 Mand 0.050 M, respectively.
3) Draw a picture of the galvanic cell represented by the line notation below and then write the reduction half-reactions for each electrode. Also identify the anode and cathode for each half cell. Pt (s) | Fe3+ (aq), Fe2+ (aq) || Cr2O72-(aq), Cr3+ (aq), HA (aq) | Pt (s)
Gibbs free energy= -434250 J/mol for a galvanic cell whose two half-reactions: Mt3+ (aq) + 3 e1- → Mt (s), and R2+ (aq) + 2 e1- → R (s), have standard reduction potentials of 1.09 V and 0.34 V respectively