Standard Electrode Potential + Feasability Flashcards
Standard electrode potential
The potential difference of a cell measured where this half cell is connected to the standard hydrogen electrode under standard conditions
Negative sign on a standard electrode potential value
Shows that when connected to the standard hydrogen electrode potential, it is oxidised and loses electrons to be given to the SHE
ACTS AS REDUCING AGENT
Positive sign on a standard electrode potential
Shows that when connected to the standard hydrogen electrode potential, it is reduced and gains electrons to be taken away to the SHE
ACTS AS OXIDISNG AGENT
How do we know what half cell is oxidised?
If it’s more negative
E cell
A calculated value by adding the standard electrode potential values
How to calculate the Ecell value? Flipping method
Oxidised half cell has the sign is flipped
Reduced half cell is kept the same
If Ecell value is positive?
The reaction is THERMODYNAMICALLY feasible UNDER STANDARD CONDITIONS
So could possible happen spontaneously
If Ecell is negative?
The reaction is THERMODYNAMICALLY NOT feasible UNDER STANDARD CONDITIONS
So could not possibly happen under standard conditions
What is the standard hydrogen electrode?
We use as a standard to measure the potential difference of half cells because its voltage is 0
So any potential difference is equal to the E value of the half cell
What set up is is the standard hydrogen electrode?
Pt electrode
System to insert H2 gas in
Solution containing 1 molddm-3 H+ ions
What set up is is the standard hydrogen electrode?
Pt electrode
System to insert H2 gas in
Solution containing 1 molddm-3 H+ ions
Equation for standard hydrogen electrode
2H+ +2e- <——> H2
Comparing standard electrode potentials
Comparing relative positions of equilibrium aka how likely it is to lose electrons (lies to the left) or gain electrons (lies to the right) but by using a standard hydrogen electrode to calculate these values in the first place
The potential difference of an electrochemical cell is equal to…
The Ecell of the cell calculated using standard hydrogen electrode potential values for both half cells
What does it mean if a reaction is feasible?
It has the possibility to occur spontaneously but does not necessarily mean it WILL occur
What is the equation that links e cell to entropy?
𝜟G = -nFEcell
Gibbs free energy = -(number of moles of ELECTRONS x Faradays constant x Ecell calculated)
What is n equal to in 𝜟G = -nFEcell ?
NUMBER OF MOLES OF ELECTRONS TRANSFERRED PER 1 MOLE OF REACTANT
How is E cell directly proportional to the entropy change of a reaction?
Because 𝜟G = -nFEcell
And 𝜟G = -T𝜟Stotal
So -T𝜟Stotal = -nFEcell
Divide by -Temperature gives 𝜟Stotal = nFEcell/Temperature
Because Faraday, number of moles and temperature is constant, the entropy change is therefore proportional to Ecell
What are the limits in predicting feasibility of a reaction?
Does not necessarily mean the reaction will occur if not kinetically favourable eg, the activation energy is too high
Only a prediction under standard conditions (as Ecell is measured under standard conditions) so does not take into account increasing temp
Why do we want to find the standard electrode potentials in the first place?
Because we can use these values to calculate whether a reaction (NOT NECESSARILY IN A HALF CELL) is thermodynamically feasible thus could happen (under standard conditions)
Ie find out whether less reactive metal could reduce more reactive out of solution = not feasible
Increasing the concentration (above standard conditions) of a reactant effect on half equations equilibrium
Will push the position of equilibrium to the right
So more reduction occurs thus the value for ELECTRODE POTENTIAL (no longer standard) will become more positive
This could mean a reaction could now be feasible under non standard conditions or less feasible
Increasing the concentration of a product (above standard conditions) effect on the position of equilibrium?
Will push the position of equilibrium to the left
So more oxidation occurs thus the value for ELECTRODE POTENTIAL (no longer standard) will become more negative
This could mean a reaction could now be feasible under non standard conditions or be less feasible etc
Finding if a disproportionation reaction is thermodynamically feasible
Write out both oxidation and reduction for the same element
Flipping method: flip standard electrode potential sign for oxidation one then add to reduction electrode potential, if answer is positive = thermodynamically feasible under standard conditions
How is E cell directly proportional to lnK?
𝜟Stotal = nFEcell/T but 𝜟Stotal = RlnK
So Rlnk = nFEcell/T
Divide by gas constant R
lnK = nFEcell/RT
Gas constant, number of moles of electrons, faradays are all constant at constant temp so E cell is directly proportional to lnK
