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:
B. The energy absorbed in the first move is greater than the energy released in the second move.
Explanation:
It takes large amounts of energy for a electron to jump energy levels and the further it moves, the more it takes.
Molar Mass of Calcium carbonate:-
Now
Mass = mr x moles
Mr of CuCl2 = ( 63.5) + ( 35.5 x 2) = 134.5
2.5 = 134.5 x moles
2.5 / 134.5 = moles
Moles = 0.019 (2DP)
0.25g of Al
Mr of Al = 27
0.25 = 27 x moles
0.25/ 27 = 0.0093 moles (2sf)
Hope this helps :)
Answer:
3 electrons
Explanation:
aluminum : [Ne]3s23p1 [ N e ] 3 s 2 3 p 1 . It loses 3 electrons from 3s and 3p orbitals and attains the noble gas configuration of Neon.
