Chapter Chosen


Book Chosen

Chemistry I

Subject Chosen


Book Store

Download books and chapters from book store.
Currently only available for.
CBSE Gujarat Board Haryana Board

Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for.
Class 10 Class 12
Using the standard electrode potentials given in the table 3.1(in NCERT), predict if the reaction between the following is feasible:
(i) Fe3+ (aq) and I (aq)
(ii) Ag+ (aq) and Cu(s)
(iii) Fe3+ (aq) and Br(aq)
(iv) Ag(s) and Fe3+(aq)
(v) Br2(aq) and Fe2+(aq)

From the table, standard electrode potents at 298 k are:
(E°Fe3+/VF = 0.77 V, E°I2/I– = 0.54 V)
(E°Ag+/Ag = E°Cu2+/Cu = 0.34)
(E°Fe3+/Fe= 0.77 V, E°Br2/Br- = 1.08 V)
(E°Ag+/Ag= 0.8 V E°Fe3+/Fe2+ = 0.77 V)
(E°Fe3+/Fe2+= 0.77 V,E°Br2/Br- = 1.08 V)

For a feasible reaction
rGθ  < 0
And  ∆rGθ = – nFEocell   so that
– nFEocell  <  0
n and F both are always positive values 
so that
–Eocell  <  0
Change the sign we get
Eocell     >  0
Hence for any feasible reaction Eocell will always positive

(a) Fe3+(aq) + I-(aq)  Fe2+(aq) + 12 I2 (g)Fe3+(aq) + I-(aq)  Fe2+(aq)+12I2(g)
In this reaction, Fe3+ is reduced to Fe2+ and I is oxidised to I2. The cell giving above reaction will be

I2(s) | I- (aq) || Fe3+ (aq) | Fe2+(aq) E0cell = E0cathode - E0anode = E0right - E0left               = 0.77 V - 0.54 V = + 0.23 V

As E
0 is positive, the reaction between Fe3+ (aq) and I (aq) occurs as indicated by possible reaction given above.

(b) 2 Ag+(aq) + Cu(s)  2 Ag(s) + Cu2+ (aq)
Here, in this reaction, Ag
+ is reduced to Ag (i.e., it should be cathode) and Cu(s) is oxidised to Cu2+(aq) (i.e., it should be anode).
The cell can be represented as
Cu(s) | Cu2+(aq)|| Ag+ (aq) | Ag(s)                      anode              cathodeE0cell  = E0cathode - E0anode            = 0.80 V - 0.34 V = + 0.46 V

As E°
cell is positive, the reaction between (Ag(aq) and Cu(s) occurs as indicated by possible reaction given above.

(c) Fe3+(aq) + Br-(aq)  Fe2+(aq) + 12Br2(aq)
 In this reaction Fe3+ is reduced to Fe2+ (i.e., Fe3/Fe2+ electrode should be cathode) and Br is oxidised to Br2 (i.e., Br2/Br electrode should be anode.
The cell can be represented as:

Br2(aq)|Br-(aq)||Fe2+(aq) | Fe3+(aq)E0cell = E0cathode - E0anode = E0right -E0left           = 0.77 V - 1.08 V = -0.31 V

As E°
cell is negative, no reaction will occur between Fe3+ (aq) and Br(aq).

(d) Ag(s) + Fe3+ (aq)  Ag+(aq) + Fe2+(aq)     
Two half-cell reactions can be expressed as:

Ag(s)  Ag+(aq) + e-                                    (oxidation anode)Fe3+(aq) + e-  Fe2+(aq)                                    (reduction, cathode)E0cell = E0cathode - E0anode           = 0.77 V - 0.80 V = -0.3 V

As E°
cell is negative, no reaction occurs between Fe3+(aq) and Ag(s).

(e) 12Br2(aq) + Fe2+(aq)  Fe3+ (aq) + Br- (aq)
The two half-cell reactions are

   12Br2(aq) + e-  Br- (aq)                              (reduction, cathode)      Fe2+(aq)  Fe3+(aq) + e-                                          (oxidation anode)E0cell = E0cathode - E0anode           = 1.80 V - 0.77 V = +0.31 V

As E°
cell is positive, the reaction is feasible, i.e., reaction between Br2(aq) and Fe2+ (aq) occurs as indicated by possible reaction given above.


Can you store copper sulphate solutions in a Zinc pot?


No. Because zinc is more reactive than copper and thus holes will be developed in zinc pot.

Cu2+(aq) + Zn(s) → Zn2+ (aq) + Cu(s)

Calculate the emf of the cell in which the following reaction takes place:
Ni(s) + 2Ag+ (0.002 M)  Ni2+ (0.160 M) + 2Ag(s)
Given that EoCell = 1.05 V


Ecell = Eocell + RT2FlnAg+(aq)2Ni2+ (aq)
or      Ecell = Eocell+0.0592 logAg+(aq)2Ni2+ (aq)
The equation is also written as

or    Ecell = Eocell-0.0592logNi2+(aq)Ag+ (aq)2         = 1.05 V - 0.0295 log 0.1600.002  

= 1.05 V – 0.0295 x log 80

= 1.05 V – 0.0295 x 1.9031

= 1.05 V – 0.056 = 0.99 V.


Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.


For hydrogen electrode, ,
given that      pH = 10
use formula [H+] = 10–pH
so that  [H+] = 10−10 M
Electrode reaction will
H+  + e –  →1/2 H2
Use the formula

EH+/H2  = EoH+/H2 - RTnFlnH2[H+]2              = 0-8.314×2982×96500ln IH+2              = 0.05915 log [H+]              =-0.05915 pH             =-0.05915 × 10 =- 0.59 V

Consult the table of standard electrode potentials and suggest three substance that can oxidize ferrous ions under suitable conditions.


oxidation of ferrous ion means :

Fe2+--> Fe3+ +e-EFe3+Fe2+0 = 0.77V
Any substance which standard electrode potential is more than that of Fe+3 /F+2 can oxidise ferrous ions. 
(refer to the table given in book)

The EMF of the substance whose reduction potentials greater than 0.77V will oxidised ferrous ion.
for example Br2, Cl2,and F2 .


How would you determine the standard electrode potential of the system Mg2+/Mg?


Use standard hydrogen electrode as anode and Mg2+ | Mg as a cathode we can measure the standard electrodepotential of systemMg2+ | Mg. Standard hydrogen electrode, represented by Pt(s), H2(g) (1 atm) | H+ (aq) and dip the electrode of Magnesium wire in a 1M MgSO4 solution .The standard hydrogen electrode is always zero.
Use formula
Eocell = Eo right  – Eoleft
The standard hydrogen electrode is always zero.
So that the value of
Eoleft =0
Eocell = Eo Mg|Mg2+
Eo Mg|Mg2+= Eocell