Answer:
The remaining electron will go in the 2p orbital. Therefore the B electron configuration will be 1s22s22p1.
Ecell = E°cell - RT/vF * lnQ
R is the gas constant: 8.3145 J/Kmol
T is the temperature in kelvin: 273.15K = 0°C, 25°C = 298.15K
v is the amount of electrons, which in your example seems to be six (I'm not totally sure)
F is the Faradays constant: 96485 J/Vmol (not sure about the mol)
Q is the concentration of products divided by the concentration of reactants, in which we ignore pure solids and liquids: [Mg2+]³ / [Fe3+]²
Standard conditions is 1 mol, at 298.15K and 1 atm
To find E°cell, you have to look up the reduction potensials of Fe3+ and Mg2+, and solve like this:
E°cell = cathode - anode
Cathode is where the reduction happens, so that would be the element that recieves electrons. Anode is where the oxidation happens, so that would be the element that donates electrons. In your example Fe3+ recieves electrons, and should be considered as cathode in the equation above.
When you have found E°cell, you can just solve with the numbers I gave you.
Answer is: 13181,7 kJ of energy <span>is released when 10.5 moles of acetylene is burned.
</span>Balanced chemical reaction: C₂H₂ + 5/2O₂ → 2CO₂ + H₂O.
<span>ΔHrxn = sum of
ΔHf (products of reaction) - sum of ΔHf (reactants).</span><span>
Or ΔHrxn = ∑ΔHf (products of reaction)
- ∑ΔHf (reactants).
ΔHrxn - enthalpy change of chemical reaction.
<span>ΔHf - enthalpy of formation of reactants or
products.
</span></span>ΔHrxn = (2·(-393,5) + (-241,8)) - 226,6 · kJ/mol.
ΔHrxn = -1255,4 kJ/mol.
Make proportion: 1 mol (C₂H₂) : -1255,4 kJ = 10,5 mol(C₂H₂) : Q.
Q = 13181,7 kJ.
Solid! The molecules in solid matter are arranged closely together and packed quite tightly to maintain a regular shape. Hope this helps!
