QUESTION 4Which of the following metals can be oxidized by H under standard conditions?Cu. Zn, Ag, SnAg (ag) + e Ag(S)Cu2*(aq) + 2e Cu(S)Cu2*(ag) - e → Cu*(aq)4+(aq) - 2eE = 0.80 VE° = 0.34 V%3DE = 0.15 VSnSn2+(ag) - 2e - Sn(S)Zn2*(aq) – 2eSn2+E = 0.13 VE =-0.14 VZn(S)E =-0.76 VO Zn and SnO Ag. Sn, and CuO Zn onlyAg only

Given reduction potentials are : Which of the following metals can oxidized by H+ ion = ? Options…

QUESTION 4 Which of the following metals can be oxidized by H* under standard conditions? Cu, Zn, Ag, Sn Ag (aq) - e Ag(S) Cu2*(aq) + 2e Cu(S) Cu2*(ag) - e → Cu*(aq) 4*(aq) 2e E = 0.80 V E = 0.34 V E = 0.15 V S- Sn2*(aq) Sn2*(aq)- 2e Sn(S) Zn- - Zn(S) E = 0.13 V E = -0.14 V (aq) - 2e E =-0.76 V Zn and Sn O Ag, Sn, and Cu Zn only O Ag only

QUESTION 4 Which of the following metals can be oxidized by H* under standard conditions? Cu, Zn, Ag, Sn Ag (aq) - e Ag(S) Cu2(aq) + 2e Cu(S) Cu2(ag) - e → Cu(aq) 4(aq) 2e E = 0.80 V E = 0.34 V E = 0.15 V S- Sn2(aq) Sn2(aq)- 2e Sn(S) Zn- - Zn(S) E = 0.13 V E = -0.14 V (aq) - 2e E =-0.76 V Zn and Sn O Ag, Sn, and Cu Zn only O Ag only

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

Use standard reduction potentials to find the equilibrium constant for the following reaction at 25°C: Mg?*(aq) + Pb(s) =Mg(s) + Pb2+(aq) Enter your answer in scientific notation to one significant digit. K = × 10
Balance the following reactions using the half reaction method: Al(s) + MnO4-1(aq) D Al(OH)4-1(aq) + MnO2(s) [basic solution]
The following reactions were observed: A*(aq) + B(s) → B*(aq) + A(s) A*(aq) + C(s) –→ No reaction ct(aq) + B(s) → B*(aq) + C(s) v and the strongest oxidizing Based on these observations, the strongest reducing agent is agent is 5 A(s) A(s) B(s) B(s) C(s) la C(s) A+(aq) A+(aq) + B+(aq) + B+(aq) > C+(aq) > C+(aq)
What is the total number of moles ( n) of electrons exchanged between the oxidizing agent and the reducing agent in the reaction below to obtain vanadium metal? V 2O 5(s) + 5Ca(l) → 2V(l) + 5CaO(s) 1 10 0 5 2
Al + Cu^2+ —> Al^3+ + Cu Standard reduction potential for Cu^2+ / Cu is +0.16 V, and for Al^3+ / Al is -1.70 V. Show oxidation number for each reactant and product
1. Find the standard reduction potentials for the following half-reactions in Appendix J. Cr2* (aq) + 2 e→ Cr (s) E° = V Pd2* (aq) + 2 e- Pd (s) E°
Balance the following redox reaction using the half-reaction method for acidic conditions: ?2?2(??) + MnO4 - (??) → Mn 2+(??) + O2(?)
The following skeletal oxidation-reduction reaction occurs under acidic conditions. Write the balanced OXIDATION half reaction. Mn2+ + Cr»Cr* + Mn
A solid sample containing some Fe2+ weighs 2.360 g. It required 36.44 mL 0.0244 M KMnO4 to titrate the Fe2+ in the dissolved sample to a pink end point. The balanced redox reaction is shown below. 8 H+ + MnO4ˉ + 5 Fe2+ ® Mn2+ + 5 Fe3+ + 4 H2O Calculate each of the following quantities. How many moles MnO4– were used in the titration? How many moles Fe 2+ are there in the sample? How many grams of Fe are there in the sample? What is the percentage of Fe (by mass) in the sample? Calculate the percentage of Fe in pure iron(II) ammonium sulfate hexahydrate: (NH4)2[Fe(SO4)2]• 6 H2O.
/variants/1189727/take/9/ ☆ 88 MULTIPLE CHOICE Question 10 Which half-reaction correctly represents oxidation? Fe(s) → Fe +2 A Fe(s) B Fe +2 (aq) + 2e¯ (aq) → Fe(s) + 2e- + 2e → Fe+2 (ag) D Fe +2 (aq) + 2e¯ → Fe(s)
Balance the following reaction in acidic solution & determine the coefficient of H+ Zn(s) + MnO4-(aq) --> Zn+2(aq) + Mn+2 (aq) Group of answer choices 2 4 8 16
Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: Cu²+ (aq) + 2Cu+ (aq) → Cu(s) + 2Cu²+ (aq) 2+ Cu²+ (aq) + 2e → Cu(s) Ee = 0.337 V red 0.153 V = 2+ Cu²+ (aq) +e → Cut (aq) E red AG° kJ K for this reaction would be than one.