ELECTROCHEMISTRY Flashcards
1
Q
galvanic cell converts what energy into what?
another name of galvanic cell
A
an electrochemical cell that converts chemical energy of a spontaneous redox reaction into electrical energy
voltaic cell
2
Q
what happens to gibbs free enegry in elevtrochemical cell?
A
the gibbs free energy of the spontaneous redox reaction is converted into electrical work
3
Q
when does electrochemical cell functions as ELECTROLYTIC CELL?
A
if an external opposite potential more than 1.1 v is applied
e- flow - cu to zn
current- zn to cu
zn is deposited and cu dissolved
4
Q
oxidation half and reduction half of Daniel cell?
A
oxidation: Zn(s) —-> Zn2+ +2e-
reduction: Cu2+ + 2e—–> Cu(s)
5
Q
standard electrode potential
A
when concentration of all species involved is unity then the electrode potential is said to be SEP
6
Q
why platinum or gold used as inert electrodes
pt(s)lH2(g, 1 bar) l H+(aq, 1M) ll M+(aq. 1M) l M
A
- they don’t participate in the reaction
- procide their surface for the conduction of e-
6
Q
SHE
Eo for SHE is 0
A
- electrode is platinum electrode coated with platinum black
- the electrode is dipped in acid and pure H2 is bubbled thru it
- the conc of both reduced and oxidised forms of H2 is kept at unity i.e, pressure = 1 BAR and conc = 1 molar
7
Q
strongest oxidising agent
hitthedance
A
FLUORINE
8
Q
weakest reducing agent
A
fluoride ions
9
Q
strongest reducing agent
hint hint: ______________aluminium hydride
A
lithium metal
9
Q
weakest oxidising agent
A
lithium ion
10
Q
what if Eext is = 1.1V
A
no flow of current or electron and no chemical reaction
11
Q
if Eext<1.1V then?
A
- e- flow from zn to cu
- current flow from cu to zn
- zn is dissolved at anode and cu is deposited at cathode
12
Q
SALT BRIDGE :what and why
A
consist of inverted U tube filled with paste of agar- agar or gelatine in aconcentration solution of an inert electrolyte like KCL, NH4NO3
functions:
* complete the electrical circuit by allowing ions to move from one solution to the other without mixing the solution
* to maintain electrical neutrality
13
Q
Ecell
the difference between electrode potential of the 2 half cell
A
Ecathode - Eanode
Eright - Eleft
Reduction potential of cathode - Reduction potential of anode
14
Q
can you store copper sulphate in a zinc pot
A
Zn displaces Cu from copper sulphate solution while Cu doesnt displace zn from zinc sulphate solution
15
Q
Li Mg &Al Zn IRoNI Tinil aaki HYDROGEN Coril AgFricayilekk pooyi
MORE -VE THE VALUE STRONGER REDUCING AGENT
MORE +VE THE VALUE STRONGER OXIDISING AGENT
A
- lithium li2+/li
- magnesium mg2+/mg
- aluminium al3+/al
- zinc zn2+/zn
- iron fe3+/fe
- tin sn2+/sn
- hydrogen 2H+/H2
- copper Cu2+/Cu
- silver Ag+/Ag
- fluorine F2/2F+
16
Q
nernst equation
A
Ecell= Eocell - 0.059/n LOG [P]^p/[R]^r
17
Q
kohlraush law
A
limiting molar conductivity of an electrolyte can be represented by the sum of contribution of anions and cations of the electrolyte
18
Q
application of Kohlraush’s law
A
- alpha = lamda m / lamda o m
- k dissociation = C alpha^2/(1-alpha)
19
Q
ELECTROLYTIC CELL
A
ELECTRICAL ENERGY TO CHEMICAL ENERGY
20
Q
Electrolysis
A
the process of decomposition of electrolyte by the passage of electricity thru the aqueos solnor molten state
21
Q
how many charge is required for the reduction of Mno4- to Mn2+
A
- oxidation no. of Mn in Mno4- is
x +(-2)4 = -1
x = 7 - to reduce Mn+7 to Mn+2 5e- is required
- therefore 1 mole Mn2+ need 5F
22
Q
A Solution of CuSo4 is electrolysed for 10 mins with 1.5 A. what is the mass of cu deposited at the cathode
A
- write the reduction half reaction
cu2+ +2e- = cu - calculate charge with e- for one mol of cu
2F= 2* 96500 = 193000C - calculate charge with current and time (sec)
Q = I x t
1.5 x 600 = 900C - calculate no. of moles needed for 900C
900c = 1 mo/193000*900
4.66 x 10^-3 mol cu - calculate mass by mutliplying 1 mole of cu with 4.44 x 10 ^-3
1 mole of cu = 63g
4.66 x 10^-3 mole of cu = 63 * 4.66 x 10^-3
= 0.296g
23
predict the products of electrolysis
an aqueous soln of AgNO3 with **silver electrodes**
| anode - oxidation
## Footnote
cathode - reduction
cathode: Ag+ +1e- ----> Ag
anode: **Ag -----> Ag+ +1e-**
| for purification of metal
## Footnote
reduction potential more reduction occurs
reduction potential less oxidation occurs
24
predict the products of electrolysis
a dilute soln H2SO4 with platinum electrodes
| anode :SO42->NO3->OH->Cl->Br->I-
## Footnote
cat: ions same
* anode: OH- --->OH + e
4OH- -------> 2H2O +O2
* cathode: H+ +e- -----> H2
H+ + H+ ---->H2 (g)
| anode ions are: SO42- &OH-
## Footnote
cathode ions are: H2 &H+
25
predict the products of electrolysis
an aqueous soln of CuCl2
| anode :SO42->NO3->OH->**Cl**->Br->I-
## Footnote
cat: Li Mg &Al Zn IRoNI Tinil aaki **H**YDROGEN **Co**ril AgFricayilekk pooyi
* anode: Cl- ------------> Cl + e-
Cl + Cl ------> Cl2(g) is liberated
* cathode: Cu2+ + 2e- -----> cu
cu is deposited
| anode: ions are 2Cl- and OH-
## Footnote
cat: ions are cu2+ and H+
26
If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?
1. find charge
I = 0.5 A t = 2 hours = 2 × 60 × 60 s = 7200 s
Thus, Q = It = 0.5 A × 7200 s = 3600 C
2. We know that 96500C = 6.02 X 10^23 number of electrons. (1 mole of e-)
3. for 3600 C => 6.02 x 10^23 / 96500 * 3600
**2.25 x 10^22 e-**
| 1 mole of e- = 6.02 x 10^23
27
Electronic Conductance
* electrical conductance thru metals
* is due to movement of e-
* depends on
1. nature and structure of metal
2. valence e- per atom
3. temp (increases conductance decreases)
* as e- moves from one end to another there is no change in composition
28
Ionic conductance
| AKA electrolytic conductance
* the conductance of electricity by ions present in the soln
* depends on:
1. nature of electrolyte
2. conc on electrolyte
3. nature of solvent and viscosity
4. size of ions produced and their solvation
5. temp (increases ionic conductance increases)
* passage of DC current for prolonged time can change composition due to electrochemical reaction
29
What is the problem faced by measuring ionic condutance and how to resolve them
1. passing DC current changes the composition of the solution
2. a soln cant be connected to the bridge like a metallic wire or other solid conductor
RESOLVED
1. passing AC current
2. use a conductivity cell
30
Conductivity cell and CELL CONSTANT
1. Consist of 2 pt electrode coated with pt black
2. l/A
G* = l/A = Rk
| SI UNIT: cm-1
31
Why do conductivity of soln of different electrolyte in the same solvent at a given temp differ?
1. charge
2. size of ions in which they dissociate
3. the conc of ions / ease with which the ions move under potential gradient
## Footnote
a potential gradient is the local rate of change of the potential with respect to displacement,
32
Molar conductivity
λm = k x 1000 / molarity
is the conductiance of the volume V of soln containing one mole of electrolyte kept btwn 2 electrodes with area of cross- section A and distance of unit length
λm = k V
| si unit: Scm^2 mol-1
33
why does conductivity for weak and strong electrolyte decrease with increase in dilution?
| or decrease in concentration
the no. of ions per unit volume that carry the current in a soln decreases on DILUTION
34
Limiting Molar Conductivity
when conc approaches 0 the molar conductivity is called LIMITING MOLAR CONDUCTIVITY
35
why MOLAR conductivity increase with decrease in conc?
| i.e, increase in dilution
the total V of soln containing 1 mole of electrolyte also increases
| V - volume
## Footnote
decrease in conductivity (k) on dilution is more than compensated by increase in V
36
**KOHLRAUSCH LAW OF INDEPENDENT MIGRATION OF IONS**
| there is A LIMIT
limiting molar conductivity of an electrolyte can be represented by the sum of individual contributions of anion and cationsof the electrolyte
37
WEAK ELECTROLYTES
1. at high conc degree of diss is lower
2. molar conductivity on dilution increases due to increase in degree of diss and consequently no. of ions per volume of soln containing 1 mole of electrolyte
3. limiting molar conductivity at low conc cant be determined by etrapolating molar conductivity to zero as it increases steeply
4. at infinite dilution electrolyte dissociate completely
38
α
λm/λm0
39
Kα
cα^2/(1-α)
40
how are metals like Na, Mg, Al produced on large scale when there no suitable reducing agent present
show example with cu
1. cathode (-vely charged): cu2+ +2e----> cu(s)
copper metal is deposited at cathode (reduced)
2. anode (+vely charged): cu(s) ---> cu2+ + 2e-
copper is dissloved (oxidised) at anode
3. this is basis of industrial process in which impure cu is converted to pure cu
4. impure cu is made an anode that dissolves and pure cu is deposited on cathode
## Footnote
sodium and magnesium are produced by electrolysis of their fused chlorides
# alumiunum by electrolysis of aluminum oxide in presence of cryolite
41
Faraday's first LAW of electrolysis
| note: here chem rnx and by current
the amount of chemical reaction😪 which occurs at any electrode during electrolysis by current is proportional to the quantity of electricity passed through the electrolyte.
42
Faraday's second LAW of electrolysis
the amount of different substances liberated by the same quantity of electricity passed through the electrolytic soln is proportional to their chemical equivalent weights
## Footnote
Weq = atomic mass/ no. of e- transferred
43
# charge required for one mole
1. Ag+ +e- = Ag(s)
2. Mg2+ + 2e- = Mg (s)
3. Al3+ +3e- = Al (s)
1. 1F
2. 2F
3. 3F
## Footnote
1mole of e- = reduce/ oxidise 1 mole of metal ion
44
WHAT does products of electrolysis depend on
1. nature of material being electrolysed
2. type 0f electode being used
## Footnote
if electrode is inert: it doesnt participate in rnx and acts as source or sink for e-
45
Consider the reaction: Cr2O72--+ 14H+ + 6e- -> 2Cr3+ + 7H2O What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr2O72- ?
1. oxidation no of Cr=
2x +(-2)7 =-2
x = 6
2. 1 mole of e- = 1F
here 6 mole of e- needed to reduce Cr2O7 2-
so **6F**
46
# PREDICT THE PRODUCTS
NaCl(aq)----------->Na+ +Cl-
anode: Cl- (aq) ---> 1/2 Cl2 (g) + e-
cathode: H2O (l) + e- -->1/2H2 (g) +OH-
net rnx: NaCl(aq) +H2O(l)--> Na+ (aq) + OH- (aq) + 1/2 H2(g) + 1/2 Cl2(g)
| note: H+ is produced by diss of H20
H20 -> 1/2 H2 + OH-
## Footnote
1. cathode= competition btwn Na+ and H+ - consider rnx with high E0cell
2. anode= comp with Cl and H20 - consider OVERPOTENTIAL instead of low E0cell
47
# PREDICT THE PRODUCTS OF
electolysis of SULPHURIC ACID preffered AT ANODE
1.DILUTE h2so4
2.CONC h2so4
1. 2H20(l) --> O2(g) + 4H+ +4e-
2. 2SO4^2- (aq) --->S2O8^2- (aq) + 2e-
48
how to determine the practicality of a battery
1. light
2. compact
3. voltage should not vary during its use
48
# name the batteris used in
1. transistors
2. clocks
3. hearing aide
4. watches
5. automobiles
6. inverters
7. cell more expensive than LEAD STORAGE
8. apollo space programme
1. lechlanche
2. drycell
3. mercury
4. mercury
5. lead storage
6. lead storage
7. nickel cadmium
8. fuel cell
49
advantagous of fuel cell
1. pollution free
2. high efficiency
3. runs continuosly
50
how to prevent CORROSION
1. PAINT e.g. bisphenol
2. cover surface with other metal (sn, zn,)
3. electrochemical methode to provide sacrificial electrodeof another metal(which corrode itself)
| sacrificial metals have reduction potential lower than iron
## Footnote
zn, al, mg- oxidation occur faster than iron
51
# primary battery
LECHLANCHE/ DRY CELL
| transistor and clock
## Footnote
rnx occur only once and after use the batter becomes dead and cant be reused
* cell potential: 1.5 V
* electrolyte: moist paste of NH4Cl and ZnCl2
* anode: Zn
* cathode: graphite
## Footnote
Mn is reduced from +4 to +3
ammonium produced form a complex with Zn2+ -> [Zn(NH3)4]2+
52
# primary battery
MERCURY CELL
| watches and hearing aides
## Footnote
SUITABLE FOR LOW CURRENT DEVICE
* cell potential: 1.35V
* electrolyte : paste of ZnO and KOH
* anode : zn mercury amalgam
* cathode: paste of HgO and carbon
## Footnote
cell potential remains contant as the overall reaction doent involve any ion in soln whose conc change during its lifetime
53
# secondary battery
LEAD STORAGE
| can be recharged & undergo large no. of discharging and charging
## Footnote
automobiles and invertors
on **Discharging**
* electrolyte : 38% soln of sulphuric acid
* anode: lead
* cathode : a grid of lead packed with PbO2
| on **charging** reaction is reversed
54
# secondary battery
NICKEL CADMIUM
| EXPENSIVE &longer life
* anode : Cd
* cathode: metal grid containing nickel oxide
55
# fuel cell
H2- O2 FUEL CELL
| 2 materials other than H2 that can be used as** fuels**
## Footnote
Methane (CH4) and methanol (CH3OH)
* 70% more efficient (thermal plant is only 40%)
* electrolyte: NaOH
* Catalyst : finely divided pt or palladium
* anode: 2h2
* cathode: o2
* electrode: carbon
| fuel cell- convert energy of combustion of fuels to electrical energy
