lecture 4 fr

Question 1

describe a galvanic cell cathode

Answer 1

• reduction occurs
• ions gain electrons to form solid
• the cathode itself is positive (e- go to it from the anode)

Question 1

what is cell potenial E,, Ecell

Answer 1

the voltage measured between the 2 electrodes

Question 2

describe a galvanic cell anode

Answer 2

• oxidation occurs
• solid becomes ions as it loses e-
• the anode itself is negative

Question 3

cell potential,, Ecell and gibbs free energy

Answer 3

Gibbs = - ve x F x Ecell

• ve = number of electrons (balanced equation)
  F = faraday constant
  Ecell = cath - anode

Question 4

if Ecell is positive,, the reaction is

Answer 4

spontaneous

Question 5

if gibbs is negative the reaction is

Answer 5

spontaneous

Question 6

G is also (kinda weird)

Answer 6

G = chem potential of products - chem potential of reactants

Question 7

meaning G is also (even weirder)

Answer 7

G = sum of: electrons x chemical potential

electrons for products = positive

electrons for reactants = negative

Question 8

how can u find the chemical potential of smt when u have the standard chemical potential

Answer 8

chemical potential = standard chemical potential + RT ln(a)

Question 9

what can we subs in for G

Answer 9

chem potential of products - chem potential of reactants

Question 10

what is ub and ua and what can we sub in for them

Answer 10

ub = chemical potential of product

ub = ub* + RTln(a)

we can sub these in when doing G= ub - ua

Question 11

gibbs when u have standard gibbs

Answer 11

G = G* + RTln(Q)

Question 12

Gibbs when we have Ecell or number of electrons or faraday constant

Answer 12

G = -vFE

Question 13

-vFE is equal to what

Answer 13

Gibbs standard + RTln(Q)

Question 14

whats is E* standard

Answer 14

• (Gibbs standard / vF)

Question 15

how can the potential difference for smt be found

Answer 15

defined in respect to a reference : H half cell

Question 16

H calf cell

Answer 16

Pt | H2(g) |H+ (aq)

Question 17

measuring the actual cell potentials

Answer 17

use H cell

E = cathode - anode

anode = 0 bc H half cell

so E = cahode E

Question 18

what does the nernst equation imply

Answer 18

any 2 cells with the same total reaction and the same number of electrons will have the same E(cell potential)

Question 19

from Cu 2+ -> Cu+

and Cu+ -> Cu

can the electrode potentials just be added to get from
Cu2+ -> Cu

Answer 19

nope!!
electrode potentials are not additive

u cant just add the Ecell

Question 20

so for elements with different oxidation states,, what should u do when u need to combine 2 equations

Answer 20

use gibbs!!
G = -vFE
multiple E byt the faraday constant and number of electrons to get Gibbs.

do this for each one

then u can add 💗

Question 21

for elements with different oxidation states when given the half reactions as reductions, and the Ecell***

Answer 21

Ecell(C) =
vA x E*A + vB x EB// vC

aka electrons from reduction of A x E*cell of A

add electrons from reduction of B x E*cell of B

then divide that by the total number of electrons in both reactions!!

the electrons here are not negative tho!!

Question 22

when combining reduction reaction using the
vEcell + vEcell // Total v

what are tips and stuff to look out for

Answer 22

when they give u the equation to find Ecell..

look at the product and the reactant and see which one of the 2 reactions given have these..

u may have to flip the equation and the sign of the Ecell in order to figure out the correct answer.

Mn 3+ -> Mn2+ (Ecell = 1.560)
Mn 3+ -> Mn (Ecell = - 0.268)

so to get from Mn2+ -> Mn

Mn2+ -> Mn3+ -> Mn

u flipped the Mn3+ -> Mn2+ equation so u must also flip the sign 💗

Question 23

gibbs and rate constant

Answer 23

G = - RTln(k)

e(G/-RT) = k

Question 24

nernst equation: Q

Answer 24

basically just K!!
don’t include the solid tho!!

just conc of ions!!