Cell: Fe(s)| Fe2+(aq, 1M) | Cu2+(aq, 1M) | Cu(S)
Calculated Values
ΔG° = -nFE° = ΔH° - TΔS° ΔH° = -0.34905 J/C * 2 mols * 96485 C/mol = -67,356.18 J =67.36kJ
ΔS° = 0.8597 x 10-3 J/C * 2 mols * 96485 C/mol = 165.89 J
ΔG° =-67356.18 J – (298*165.9 J) = -116794.4 J = -116.79 kJ
Theoretical Values ΔG⁰ = -(2 mols *96485 C/mol * 1.07 J/C) = -206.48 kJ
ΔH⁰ =
ΔS⁰ =
Cell: Zn(s)| Zn2+(aq, 1M) | Cu2+(aq, 1M) | Cu(S)
ΔG° = -nFE° = ΔH° - TΔS°
ΔH° = -0.87252 J/C * 2 mols * 96485 C/mol = -168,370.1844 J -168.4kJ
ΔS° = -0.4069 x 10-3 J/C * 2 mols * 96485 C/mol = 78.52 J
ΔG° = -168,370.54 J – (298*78.52 J) = -191768.96 J = -191.77 kJ
Theoretical Values ΔG⁰ = - (2 mols * 96485 C/mol * 1.10 J/C) = -212.267 kJ
ΔH⁰ =
ΔS⁰ =
** Can someone please help calculate the theoretical ΔS⁰ and ΔH⁰ for both equations. Please show work, thanks!
Part 1:
Theoretical ΔHo = -157.04 kJ.
ΔSo = 165.89 J.
Explanation:
The theoretical ΔGo has been obtained using the relation ΔG° = -nFE°.
The relation between ΔSo and E° is:
Hence the theoretical value for ΔSo will be this, which is given here to be 165.89 J.
Since the ΔSo value is given, we can obtain ΔHo value from the relation:
ΔHo = ΔGo(theoretical) + T ΔSo
= -206.48 + (298 * 0.16589) = -157.04 kJ
Part 2:
Theoretical ΔHo = -188.87 kJ.
ΔSo = 78.52 J.
Explanation:
Here ΔSo = 78.52 J, which will also be the theoretical ΔSo value.
Theoretical ΔHo = ΔGo(theoretical) + T ΔSo
= -212.267 kJ + (298 * 0.07852) = -188.87 kJ.
Cell: Fe(s)| Fe2+(aq, 1M) | Cu2+(aq, 1M) | Cu(S) Calculated Values ΔG° = -nFE° = ΔH°...
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