For an acidic solution:
Break the overall equation into 2 half-RXNs
Balance everything but H and O
Balance O by adding H2O as needed
Balance H by adding H+ as needed
Add electrons as needed
Multiply each half-RXN by integers to cancel the electrons
Add the 2 half-RXNs and simplify (similar in concept to Hess’s Law)
For basic solutions, you do all the above but you add enough OH- to cancel any H+ by turning it into water, and simplify again from this
The electrons would flow toward to cathode (when given a path)
They then flow spontaneously
The difference in PE and charge matters
The magnitude of the charge/# of electrons being moved does NOT matter and doesnt change voltage
SHE is the reference point for all reduction potentials, being arbitrarily assigned
2H+(aq, 1M (since its at standard conditions))+2electrons \n →H2(g, 1 atm)
Eºred=0 since its the reference point
That the reaction is spontaneous IN THAT DIRECTION
The reaction naturally wants to be reduced
You must flip the reaction to get it to be spontaneous, otherwise its non-spontaneous in that direction
The reaction is less reduced and more oxidized
F2
Cl2, Br2, I2, and any oxyanion where the central atom has a large + charge (ex: MnO4^-, Cr2O7^3-,ClO3^-, etc…. )
Strong oxidizers attract e- very strongly, their own and other substances too
A +Eº means a spontaneous reaction in that direction
A -Eº means a non-spontaneous reaction in that direction
ΔG=-nFE (or ΔG=-nFEº in standard states)
n= # of moles of transferred electrons total (after doing both half-RXNs to find its total), F=Faradays constant (96485 J/Vmol, or with units C/mol)
The Nerst equation:
The form used when at non-standard conditions:
E=Eº-((2.303RT)/(nF))logQ
The form used when at 25ºC/298k or STP:
E=Eº-((0.0592)/(n))logQ
R=gas constant (3 forms: 8.314 J/molk (most commonly used here), 0.08206 Latm/molk, and 62.36 Ltorr/molK)
n= # of moles of transferred electrons total (after doing both half-RXNs to find its total)
F=Faradays constant (96485 J/Vmol, or with units C/mol)
It will eventually equalize the two concentrations
Since both sides will eventually equalize, you’re going from a concentrated anode (giving the materials/electrons to plate the cathode) and the diluted cathode (accepting the materials/electrons to be plated), meaning that by going from a concentrated solution (R) to a dilute solution (P) makes Q=[dilute]/[concentrate] (photo of Ni included as an example to show the cancelation)
We can simply rearrange the Nerst equation to make everything equal to logK(or Q):
At non-standard conditions: logK=(nFEº)/(2.303RT)=(-ΔGº)/(2.303RT)
At standard conditions (25ºC/298k): logK=16.912(nEº)=(-ΔGº)/(5706)
Wmax=ΔG=-nFE
Wmax must be - as it must exert work onto its surroundings, not gain it
W=nFE(ext)
E(ext) simply means the external voltage source
W must be +
E does NOT equal E(ext)