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Class 10 Class 12

A Cell is prepared by dipping a copper rod in 0.01 M copper sulphate solution, and zinc rod in 0.02 M ZnSO4 solution. The standard reduction potentials of copper and zinc are + 0.34 V and – 0.76 V respectively.
(a) What will be the cell reaction?
(b) How will the cell be represented?
(c) What will be the emf. of the cell?

(a) As E°Zn2+/Zn < E°Cu2+/Cu . Zinc electrode acts as anode and copper electrode acts as a cathode. The cell reaction is
Zn(s) + Cu2+(aq) (0.01 M) → Zn2+(aq) (0.02 M) + Cu(s)
(b) The cell is represented as
Zn(s) | Zn2+(aq) (0.02 M) || Cu2+(aq) (0.01 M) + Cu(s)
(c) The emf of the cell can be calculated by using Nernst equation

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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 .

879 Views . 9 Shares

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 . 10 Shares

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)
1384 Views . 10 Shares

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

1279 Views . 3 Shares

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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