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Class 10 Class 12
Calculate the potential (emf.) of the cell
Cd | Cd2+ (0.10 M) || H+(0.20 M) | Pt, H2 (0.5 atm)
(Given : E° Cd2+ / Cd = – 0.403 V, R = 8.314 K–1 K–1 mol–1, F = 96500 C mol–1

Cell reaction will be
                            Cd+2H+    Cd2++H2                     (hence n = 2)

E°cell = E°right - E°leftE°cell = 0-(-0.403 = + 0.403 V)

Ecell  = E°cell + 2.303nFlog H+2 CdpH2Cd2+

           = 0.403+8.314 × 298 × 2.3032 × 96500log (0.2)2(0.5) (0.1)= 0.403 + 0.0295 log 0.8= 0.403 - 0.0028

Ecell = 0.400 V

Equilibrium Constant from Nernst Equation: If the Daniell cell is short circuited, then we note that the reacton.
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
takes place and as time passes, the concentration of Zn2+ keeps on increasing while the concentration of Cu2+ keeps decreasing. At the same time, voltage of the cell as read on the voltmeter keeps on decreasing. After some time, we shall note that there is no change in the concentration of Cu2+ and Zn2+ ions and at the same time, voltmeter gives zero reading. This indicates that equilibrium has been attained. In this situation the Nernst equation may be written as
Ecell° =2.303RTnFlog [Cu2+][Zn2+]Ecell° =2.303RT2Flog [Cu2+][Zn2+]
But at equilibrium [Zn2+] / [Cu2+] = K
and the above equation can be written asEcell° =0.05912logK
 (E° = 1.1 V)
K = 2 x 1037 at 298
In general E°cell = 2.303 RT / nF x log K


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 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 .


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.


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