Compute the standard cell potential for the spontaneous reaction (ie. where Sr(s) is being oxidized) at 25 °C. O 3.24 V O 8.02 V O 2.56 V O 3.24 V

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Chapter17: Electrochemistry
Section: Chapter Questions
Problem 29E: For the standard cell potentials given here, determine the ?G for the cell in k].. (a) 0.000V,n=2...
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If a voltaic cell is created between a Sr2+/Sr electrode and a In3+/In electrode

Compute the standard cell potential for the spontaneous reaction (ie. where Sr(s) is being oxidized) at 25 °C.
O 3.24 V
O 8.02 V
O 2.56 V
O 3.24 V
Transcribed Image Text:Compute the standard cell potential for the spontaneous reaction (ie. where Sr(s) is being oxidized) at 25 °C. O 3.24 V O 8.02 V O 2.56 V O 3.24 V
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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