ï»¿ What are the products obtained during electrolysis of CuSO4 using Cu electrodes? from Chemistry Electrochemistry Class 12 Nagaland Board

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What are the products obtained during electrolysis of CuSO4 using Cu electrodes?

Answer:

The electrolysis of an aqueous solution of copper sulphate using copper electrodes (i.e. using active electrodes) results in transfer of copper metal from the anode to the cathode during electrolysis. The copper sulphate is ionised in aqueous solution.

CuSO4 - -> Cu2+ + S${O}_{4}^{2-}$

The positively charged copper ions migrate to the cathode, where each gains two electrons to become copper atoms that are deposited on the cathode.

Cathode: Cu2+ + 2e → Cu

At the anode, each copper atom loses two electrons to become copper ions, which go into solution.

Anode: Cu → Cu2+ + 2e

The sulphate ion does not take part in the reaction and the concentration of the copper sulphate in solution does not change. The reaction is completed when the anode is completely eaten away. This process is used in electroplating

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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

1279 Views

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 .

879 Views

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
2203 Views

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.

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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)
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