Here all the carbon of K2CO3 converted into a complex hence POAC can apply
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Sunday, May 9, 2021
27.6 gm K2CO3 (138) was treated by a series of reagents so as to convert all its carbon into K2Zn3[Fe(CN)6]2(MM=698) Calculate mass of K2Zn3[Fe(CN)6]2.
Friday, May 7, 2021
Metal-Metal- Insoluble salts electrode:
These consist of a metal (M) which is covered by a layer of sparingly soluble salt (MX), immersed in solution containing common ions (X-). And it is represented as anode or anion electrode.
M I MX (sat’d)
I X- for example Ag I AgCl (sat’d) I Cl- II
Cathode
We Know that
AgX (F, Cl, Br,I) Silver salt’s are insoluble in water.
Other examples:
(1) Hg(s) I
Hg2SO4(sat’d) I SO4-2 II (Cathode)
used in the standard Weston Cell.
(2) Hg(s) I
Hg2Cl2(sat’d) I Cl- (KCl paste) II (Cathode)
used as Calomel electrode
(3) Pb(s) I
PbSO4(sat’d) I SO4-2 II
(Cathode) used in lead batteries
(4) Cu(s) I CuCl2(sat’d)
I Cl- (KCl paste) II (Cathode)
(1) Silver electrode as Anode or anionic electrode:
Ag I AgCl (sat’d)
I Cl- II Cathode
The
electrode reaction can be considered as taking place in two steps as.
This electrode
reaction involve only the concentration of chloride ion, the electrode is said
to be reversible with respect to chloride ions (Cl-).
The above
reaction show how an electron can be released or taken up in such electrode.
Nernst equation for this electrode (Anode) is
written as:
Silver electrode as cathode or cationic electrode:
Anode
II Cl- aq I AgCl(sat’d)IAg(s)
Nernst equation for this electrode (Cathode) is written as:
(2) Calomel electrode:
A calomel
cell consist of a platinum electrode dipping into mercury in contact with
calomel (dimercury (I) chloride,Hg2Cl2 and potassium
chloride solution.
As Anode:
Hg(s) I Hg2Cl2(sat’d)
I Cl- (KCl paste) II (Cathode) :
As Cathode:
Anode II Cl-
(KCl paste)I Hg2Cl2(sat’d) I Hg(s)
Two weak acid solution HA1 and HA2 each with same concentration and having Pka values 3 and 5 are placed in contact with hydrogen electrode ( 1 atm at 25 degree) and are interconnected through a salt bridge find the emf.
Consider the cell:
Pt H2(1atm)IHA2(1M) II HA1(1M)IH2(1 atm)Pt
Given that PKa1 of HA1 is = 3 (at Cathode)
Pka2
for HA2 is = 5 (at anode)
A current of 3 ampere passed for 2 hours through a solution of CuSO4 3 gm of Cu2+ ions were discharged at cathode calculate the current efficiency.