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If a voltaic cell is created between a Sr2+/Sr electrode and a In3+/ln electrode, what species is being oxidized? O In3+(aq) O s(s) O s2*(aq) O In(s)

If a voltaic cell is created between a Sr2+/Sr electrode and a In3+/ln electrode, what species is being oxidized? O In3+(aq) O s(s) O s2*(aq) O In(s)

Chemistry
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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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If a voltaic cell is created between a Sr2+/Sr electrode and a In3+/ln electrode, what species is being oxidized? O In3 (aq) O Sr(s) O s2*(aq) O In(s)
Transcribed Image Text:If a voltaic cell is created between a Sr2+/Sr electrode and a In3+/ln electrode, what species is being oxidized? O In3 (aq) O Sr(s) O s2*(aq) O In(s)
Consider the following table of standard electrode potentials Standard Electrode Potentials at 25 °C Reduction Half-Reaction E° (V) Au" (aq) + 3e Ag (aq) + e Fe" (ag) + 3e » Au(s) 1.50 Ag(s) 0.80 → Fe(s) → Sn(s) -0.04 Sn** (aq) + 2e -0.14 V*(aq) + 3e » V(s) -0.26 In" (aq) + 3e Eu* (aq) + 3e → Eu(s) → In(s) - 0.34 - 0.35 Cr* (aq) + 3e Та" (aq) + Зе Zn" (aq) + 2e Nb" (aq) + 3e Ti* (aq) + 2e Al* (aq) + 3e » Cr(s) -0.42 » Ta(s) -0.60 → Zn(s) → Nb(s) -0.76 -1.10 → Ti(s) -1.63 » Al(s) -1.66 Be* (aq) + 2e Am* (aq) + 2e Pr* (aq) + 2e Mg*(aq) + 2e → Be(s) → Am(s) -1.85 -1.90 » Pr(s) - 2.00 → Mg(s) → Na(s) -2.37 Na (aq) + -2.71 Ca (aq) + 2e Sr* (aq) + 2e Ва (аq) + 2e - 2.87 - 2.90 → Ca(s) » Sr(s) → Ba(s) -2.91
Transcribed Image Text:Consider the following table of standard electrode potentials Standard Electrode Potentials at 25 °C Reduction Half-Reaction E° (V) Au" (aq) + 3e Ag (aq) + e Fe" (ag) + 3e » Au(s) 1.50 Ag(s) 0.80 → Fe(s) → Sn(s) -0.04 Sn** (aq) + 2e -0.14 V*(aq) + 3e » V(s) -0.26 In" (aq) + 3e Eu* (aq) + 3e → Eu(s) → In(s) - 0.34 - 0.35 Cr* (aq) + 3e Та" (aq) + Зе Zn" (aq) + 2e Nb" (aq) + 3e Ti* (aq) + 2e Al* (aq) + 3e » Cr(s) -0.42 » Ta(s) -0.60 → Zn(s) → Nb(s) -0.76 -1.10 → Ti(s) -1.63 » Al(s) -1.66 Be* (aq) + 2e Am* (aq) + 2e Pr* (aq) + 2e Mg*(aq) + 2e → Be(s) → Am(s) -1.85 -1.90 » Pr(s) - 2.00 → Mg(s) → Na(s) -2.37 Na (aq) + -2.71 Ca (aq) + 2e Sr* (aq) + 2e Ва (аq) + 2e - 2.87 - 2.90 → Ca(s) » Sr(s) → Ba(s) -2.91
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