P2O5 = Phosphorus pentoxide
CuO = Copper (II) oxide
NH4CI = Ammonium Chloride
Mn(OH)2 = Pyrochroite
H2O2 = Hydrogen peroxide
P4S9 = Tetraphosphorus nonasulfide
CIO2 = Chlorine dioxide
NaF = Sodium fluoride
FeSO3 = Iron (II) Sulfite
Fe(NO3)3 = Iron (III) Nitrate
Cr(NO2)3 = Chromium (III) Nitrite
NaHCO3 = Sodium Hydrogen Carbonate
H2PO4 = Dihydrogen Phosphate Ion
NaCN = Sodium Cyanide
IF7 = Iodine Heptafluoride
PCI3 = Phosphorus Trichloride
In Silver, the 4d orbitals will be completely filled. That
implies that it does not have two electrons in the 5s orbital. The electronic
configuration of Silver is :
Ag (47) = 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^1
4d^10
Only the amount of gas is held constant.
Answer:
A) ΔG° = -3,80x10⁵ kJ
B) E° = 2,85V
Explanation:
A) It is possible to answer this problem using the standard ΔG's of formation. For the reaction:
Mg(s) + Fe²⁺(aq) → Mg²⁺(aq) + Fe(s)
The ΔG° of reaction is:
ΔG° = ΔGFe(s) + ΔGMg²⁺(aq) - (ΔGFe²⁺(aq) + ΔGMg(s) <em>(1)</em>
Where:
ΔGFe(s): 0kJ
ΔGMg²⁺(aq): -458,8 kJ
ΔGFe²⁺(aq): -78,9 kJ
ΔGMg(s): 0kJ
Replacing in (1):
ΔG° = 0kJ -458,8kJ - (-78,9kJ + okJ)
<em>ΔG° = -3,80x10² kJ ≡ -3,80x10⁵ kJ</em>
B) For the reaction:
X(s) + 2Y⁺(aq) → X²⁺(aq) + 2Y(s)
ΔG° = ΔH° - (T×ΔS°)
ΔG° = -629000J - (298,15K×-263J/K)
ΔG° = -550587J
As ΔG° = - n×F×E⁰
Where n are electrons involved in the reaction (<em>2mol</em>), F is faraday constant (<em>96485 J/Vmol</em>) And E° is the standard cell potential
Replacing:
-550587J = - 2mol×96485J/Vmol×E⁰
<em>E° = 2,85V</em>
I hope it helps!
