1. Refer to Table 10.1, calculate the standard-stage voltage for the following reduction half-reactions:a. Cr³+ + 3 e → Cr (s)b. MnO4 + 8 H+ + 7 e¯ → Mn (s) + 4H₂O+c. VO₂ + 4H+ +5e¯ →V (s) + 2 H₂OLet me show you an example for a. The reaction is the sum of the two reduction half-reactions:Cr³+ + e¯ → Cr²+ (Ered= -0.41 V)Cr³+ + 2 e¯ → Cr (s) (Ered = -0.91 V)Then, E=[(1)(−0.41)+ (2)(-0.91)]3= -0.74 V Reduction Half-ReactionAg+eAg(s)AgCl(s)+eAg(s) + CIAgBr(s)+ e →Ag(s) + BrAgl(s)+eAg(s) + IAg:O(s) + H₂O+ 2e →2Ag(s) + 2OH-ABS (s) + 2e →2Ag (s) +5²-Cd²+2e →Cd(s)Ce+e→Ce ³+(1 M H₂SO4)+14H* +6e2Cr³+ + 7H₂OCr³+e→Cr²+/Cr₂0₂² +Cr²+2e → Cr(s)Co³ + e→Co²+Co²+ + 2e → Co (s)Cu²+ + 2e → Cu(s)Cu²+ + e → Cu*Fe³+e→Fe²+Fe³+e-Fe²+Fete →Fe²+Fe²+2e →Fe(s)2H+2e → H₂(g)Hg²+2e → Hg (1)2+Hg²+ + 2e → Hg₂+(1 M H₂SO4)(1 M HCIO4),Eº, Vle 10-1+0.800+0.222+0.073-0.151+0.342-0.71-0.403+1.44+1.33-0.41-0:91+1.842-0.277+0.337+0.153+0.771+0.68+0.767-0.4400.000+0.854+0.919Reduction Half-ReactionMn²+ + 2e →Mn (s)MnO4+MnO²MnO4 + 4H+3e →MnO₂ (s) + 2H₂OMnO +8H + Se →Mn² + 4H₂OPb²+ + 2e →Pb(s)PbSO4(s)+2e →Pb(s) + SO.²Sn²+ + 2e →Sn (s)Sn +2e →Sn²+Ti³++eTi²+T1+2e →TI-T1+2e →TI(1 M HCIO.)TI+ eTl(s)UO₂+ + 4H+2e →U* + 2H₂OUO₂ + 4H+ e→U + 2H₂OU4+ +e →U³+VO₂ + 2H+ e → VO²+ + H₂OVO2+ + 2H+ e→V³+ + H₂OV³+ + €¯→ √²+V2+ + 2e →V (s)Zn²+2e → Zn (s)Zn(NH3)42 +2e → Zn (s) + 4NH3„E, V-1.18+0.564+1.695+1.51-0.126-0.356-0.136+0.154-0.368+1.25+1.26-0.336+0.334+0.55-0.61+1.000+0.361-0.255-1.18-0.763-1.04

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1. Refer to Table 10.1, calculate the standard-stage voltage for the following reduction half- reactions: a. Cr3+ + 3 e → Cr (s) b. MnO4 + 8 H+ + 7 e → Mn (s) + 4H₂O c. VO₂+ + 4H+ +5 e¯ → V (s) + 2 H₂O Let me show you an example for a. The reaction is the sum of the two reduction half-reactions: Cr³+ + e→Cr²+ (Ered = -0.41 V) Cr³+ + 2e → Cr (s) (Ered= -0.91 V) = -0.74 V Then, E = [(1)(-0.41)+(2)(-0.91)] 3

1. Refer to Table 10.1, calculate the standard-stage voltage for the following reduction half- reactions: a. Cr3+ + 3 e → Cr (s) b. MnO4 + 8 H+ + 7 e → Mn (s) + 4H₂O c. VO₂+ + 4H+ +5 e¯ → V (s) + 2 H₂O Let me show you an example for a. The reaction is the sum of the two reduction half-reactions: Cr³+ + e→Cr²+ (Ered = -0.41 V) Cr³+ + 2e → Cr (s) (Ered= -0.91 V) = -0.74 V Then, E = [(1)(-0.41)+(2)(-0.91)] 3

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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1a
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