Electrochemistry

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CBSE Gujarat Board Haryana Board

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Class 10 Class 12
For the cell Zn/Zn2+ (aq) || Cu2+(aq) | Cu, derive the relation between E°cell and Kc at 298 k.

Gibbs energy of the reaction given by:
${∆}_{r}G$ = – nFE(cell)
thus the reaction
Zn(s) + Cu2+(aq)---> Zn2+(aq) + Cu(s)

${∆}_{r}G$ = – 2FE(cell)

but when we write the reaction
2 Zn (s) + 2 Cu2+----->2 Zn2+(aq) + 2Cu(s)

${∆}_{r}G$ = – 4FE(cell)

If the concentration of all the reacting species is unity, then
E(cell)${\mathrm{E}}_{\mathrm{cell}}^{0}$
and we have
${∆}_{r}G$ = – nF${\mathrm{E}}_{\mathrm{cell}}^{0}$

Thus, from the measurement of ${\mathrm{E}}_{\mathrm{cell}}^{0}$ we can obtain an important thermodynamic quantity, ${∆}_{r}G$, standard Gibbs energy of the reaction.
From the latter we can calculate equilibrium constant by the equation:

${∆}_{r}G$ = –RT ln Kc

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)

Calculate the emf of the cell in which the following reaction takes place:
Ni(s) + 2Ag+ (0.002 M) $\to$ Ni2+ (0.160 M) + 2Ag(s)
Given that

Answer:

$\mathrm{Ni}\left(\mathrm{s}\right)+2{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)\to {\mathrm{Ni}}^{2+}\left(\mathrm{aq}\right)+2\mathrm{As}\left(\mathrm{s}\right)$

or

The equation is also written as

or

= 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-
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 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

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

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