BConsider the following half-reactions: Half-reaction E° (V) Hg2*(aq) + 2e" . → Hg(l) 0.855V Cu²*(aq) + 2e → Cu(s) 0.337V Mn2+ *(aq) + 2e¯ – Mn(s) -1.180V (1) The weakest oxidizing agent is: enter formula (2) The strongest reducing agent is: (3) The strongest oxidizing agent is: (4) The weakest reducing agent is:

BConsider the following half-reactions: Half-reaction E° (V) Hg2(aq) + 2e" . → Hg(l) 0.855V Cu²(aq) + 2e → Cu(s) 0.337V Mn2+ *(aq) + 2e¯ – Mn(s) -1.180V (1) The weakest oxidizing agent is: enter formula (2) The strongest reducing agent is: (3) The strongest oxidizing agent is: (4) The weakest reducing agent is:

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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A galvanic cell at a temperature of 25.0 °C is powered by the following redox reaction: 2+ 2VO (aq) + 4H* (aq) + Fe (s) → 2VO²+ (aq) + 2H₂O (1) + Fe²+ (aq) + 2+ 2+ Suppose the cell is prepared with 0.564 M VO₂ and 7.14 MH* in one half-cell and 6.21 M VO²+ and 6.29 M Fe² in the other. Calculate the cell voltage under these conditions. Round your answer to 3 significant digits. Ú
Write balanced half-reactions for the following redox reaction: 2Cu2+(aq)+Bi3+(aq)+3H2O(l)->2Cu+(aq)+BiO3-(aq)+6H+(aq) What is the reduction and oxidation?
A galvanic cell at a temperature of 25.0 °C is powered by the following redox reaction: 2VO₂ (aq) + 4H* (aq) + Fe (s) → 2VO²+ (aq) + 2H₂O (1)+Fe²+ (aq) Suppose the cell is prepared with 3.37 M VO and 3.31 MH* in one half-cell and 6.13 M VO²+ and 2.52 M Fe²+ in the other. 2+ 2+ Calculate the cell voltage under these conditions. Round your answer to 3 significant digits.
6. The above diagram shows a Galvanic constructed from Ag|AGNO3(aq) and Ni|Ni(NO3)2(aq) half-cells and the half-cell electrode potentials are as follows: Ag*(aq) + e--> Ag(s); E° = 0.80 V %3! Ni2*(aq) + 2e --> Ni(s); E° = - 0.23 V Calculate the Ecell at 25°C when the electrolyte concentrations are: [Ag*] = 1.3 x %3D 10-4 M and [Ni2+] = 0.10 M. %3D
Complete and balance the following equations for reac- tions taking place in acidic solution. (a) MnO, (aq) + H,S(aq) → Mn²+ (aq) + SO? (aq) (b) Zn(s) + NO, (aq) → Zn²* (aq)+ NH (aq) (c) H,O,(aq)+ MnO, (aq) (d) Sn(s) + NO3 (aq) (e) UO* (aq) + Te(s) → U** (aq) + TeOž¯ (aq) → 02(g) + Mn²* (aq) 4+ → Sn** (aq) + N2O(g)
R = 8.314 mol·K F = 96,485 mot AG = AG° + RT · In(Q) ΔΕΔΕ-).In(O) Half Reaction (Note: All given as reduction) E° (V) 02 (g) + 4 H*(aq) + 4 e¯ → 2 H20 (1) 1.229 2 z" (aq) Z2 (s) + 2 e 3+ (aq) + 3 е 0.426 A (s) 0.292 2 H20 (1) + 2 e G2+ (aq) + 2 e H2 (g) + 2 OH (aq) - 0.828 G (s) - 1.245 > M2+ (aq) + 2 e - 1.893 → M (s) A student constructs a voltaic electrochemical cell with two metal electrodes [metal G and metal A] in their respective aqueous nitrate solutions [G(NO3)2 and A(NO3)3]. Use this information, as well as the reduction potentials in the table above to complete each statement below. Consider the same cell from the above prompt. Calculate AG° in kJ/mol for this galvanic cell. Report your answer with 4 significant figures. You do not need to report units with your answer. If your value is negative, make sure to include a "-" symbol.
Consider the following half-reactions: Half-reaction E° (V) F2(g) + 2e → 2F*(aq) 2.870V Co2+*(aq) + 2e → Co(s) |-0.28OV 2+ Mg"(aq) + 2e → Mg(s) -2.37OV (1) The weakest oxidizing agent is: enter formula (2) The strongest reducing agent is: (3) The strongest oxidizing agent is: (4) The weakest reducing agent is: (5) Will Mg(s) reduce F2(g) to F"(aq)? (6) Which species can be reduced by Co(s)? If none, leave box blank.
A galvanic cell is constructed in which the overall reac- tion is Cr,O} (aq) + 14 H¿O*(aq) + 6 I (aq) → 2 Cr* (aq) 1 3 I,(s) | 21 H,0(€) (a) Calculate AE° for this cell. (b) At pH 0, with [C»O3¯] = 1.5 M and [I-] = 0.40 M, the cell potential is found to equal 0.87 V. Calculate the concentration of Cr³*(aq) in the cell.
A galvanic cell is constructed in which the overall reac- tion is Pb(s) + 2 H;O* (aq) → Pb²+ (aq) + H2 (g) + 2 H20(e) (a) Calculate AE° for this cell. (b) Chloride ions are added until PbCl, precipitates at the anode and [CI-] reaches 0.15 M. The cell potential is then measured to be 0.22 V at pH = 0 and PH, 1.0 atm. Calculate [Pb2+] under these conditions. (c) Calculate the solubility product constant Kp of PbCl2.
Half-reaction E0 (V) Cr3+(aq) + 3e- → Cr(s) -0.74 Fe2+(aq) + 2e- → Fe(s) -0.440 Fe3+(aq) + e- → Fe2+(aq) +0.771 Sn4+(aq) + 2e- → Sn2+(aq) +0.154 Determine the standard cell potential (E0cell) for the voltaic cell for this reaction below based the table above. Sn2+(aq) + Fe3+(aq) → Fe2+(aq) + Sn4+(aq)
(a) Following reactions occur at cathode during the electrolysis of aqueous sodium chloride solution:Na+(aq) + e-1 ——-> Na(s) E° = – 2.71 VH+(aq) + e-1 ——–> 1/2H2(s) E° = 0.00 VOn the basis of their standard reduction electrode potential (E°) values, which reaction is feasible at the cathode and why?(b) Why does the cell potential of mercury cell remain constant throughout its life?
Examine the following half-reactions and select the strongest oxidizing agent among the substances. v Jersey Departm E° = 0.755 V [PtCl4]²¯(aq) + 2e¯ 2 Pt(s) + 4CI'(aq); RuO4(s) + 8H*(aq) + 8e¯ 2 Ru(s) + 4H2O(1); FeO4²(aq) + 8H*(aq) + 3e¯ 2 Fe3*(aq) + 4H2O(/); H4XEO6(aq) + 2H*(aq) + 2e¯ 2 XeO3(aq) + 3H2O(/); E° = 2.42 V E° = 1.038 V E° = 2.07 V RuO4(s) H4XEO6(aq) [PtCl4]² (aq) O HFEO4¯ (aq) CI(aq)