ï»¿ The following chemical reaction is occuring in an electro-chemical cellMg(s) + 2Ag+(0.001 M) → Mg2+(0.10 M) + Ag(s)The E° electrode values areMg2/Mg = – 2.36 VAg+/Ag = 0.81V For the cell calculate/write:(a) (i) E° value for the electrode 2Ag+/2Ag(ii) Standard cell potential E°Cell(b) Cell potential (E)cell(c) (i) Symbolic representation of the above cell     (ii) Will the above cell reaction be spontaneous?   from Chemistry Electrochemistry Class 12 Nagaland Board

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The following chemical reaction is occuring in an electro-chemical cell
Mg(s) + 2Ag+(0.001 M) → Mg2+(0.10 M) + Ag(s)
The E° electrode values are
Mg2/Mg = – 2.36 V
Ag+/Ag = 0.81
V For the cell calculate/write:
(a) (i) E° value for the electrode 2Ag+/2Ag
(ii) Standard cell potential E°Cell
(b) Cell potential (E)cell
(c) (i) Symbolic representation of the above cell
(ii) Will the above cell reaction be spontaneous?

Mg(s) → Mg2+(aq) + 2e
2Ag+(aq) + 2e → 2Ag(s)

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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
Hence
Eocell = Eo Mg|Mg2+
Or
Eo Mg|Mg2+= Eocell
2203 Views

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-
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 .

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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

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

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.

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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

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Can you store copper sulphate solutions in a Zinc pot?