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Chemistry I

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Chemistry

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CBSE Gujarat Board Haryana Board

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Class 10 Class 12
E0Ni2+/Ni and E0Cu2+/Cu are -0.25 V and + 0.34 respectively at 298 K. Formulate the self operating galvanic cell for this electrode pair. What reaction takes place in its operation? How is the ΔG° for this reaction related to the cell e.m.f.?

The cell may be represented as:
Ni(s) +Ni2+(aq)  Cu2+ (aq)+ Cu(s)At anode : Ni(s)  Ni2+(aq) +2e-At cathode : Cu2+(aq) +2e- Cu(s)Net cell reaction :Ni(s) +Cu2+(Aq)  Ni2+ (Aq)+Cu(s)

we have given that 
E0Ni2+/Ni   =-0.25V E0Cu2+/Cu = +0.34VEcell0 =0.34-(-0.25)       = 0.59
Relationship between G  and cell e.m.f.
G = -nE°F

here n=2
F=96500 
E0cell= 0.59
plug in above equation we get 

G  =-2 x 96500 x 0.59 =-113870 Kj/mol

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

Answer:

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
Hence
Eocell = Eo Mg|Mg2+
Or
Eo Mg|Mg2+= Eocell

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

Answer:

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

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

Answer:

Ni(s)+2Ag+(aq)Ni2+(aq)+2As(s)
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.


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

Answer:


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 .


Can you store copper sulphate solutions in a Zinc pot?

Answer:

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)