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Value of standard electrode potential for the oxidation of ClX− ions is more positive than that of water, even then in the electrolysis of aqueous sodium chloride, why is ClX− oxidised at anode - Chemistry

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प्रश्न

Value of standard electrode potential for the oxidation of \[\ce{Cl-}\] ions is more positive than that of water, even then in the electrolysis of aqueous sodium chloride, why is \[\ce{Cl-}\] oxidised at anode instead of water?

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उत्तर

In the electrolysis process of sodium chloride, oxidation of water at anode requires more potential. So \[\ce{Cl-}\] is oxidized anode instead of water.

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पाठ 3: Electrochemistry - Exercises [पृष्ठ ३९]

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एनसीईआरटी एक्झांप्लर Chemistry [English] Class 12
पाठ 3 Electrochemistry
Exercises | Q III. 34. | पृष्ठ ३९

संबंधित प्रश्‍न

Calculate Ecell and ΔG for the following at 28°C :

Mg(s) + Sn2+( 0.04M ) → Mg2+( 0.06M ) + Sn(s)

cell = 2.23V. Is the reaction spontaneous ?


Calculate E°cell for the following reaction at 298 K:

2Al(s) + 3Cu+2(0.01M) → 2Al+3(0.01M) + 3Cu(s)

Given: Ecell = 1.98V


Calculate emf of the following cell at 25°C:

\[\ce{Sn/Sn^2+ (0.001 M) || H+ (0.01 M) | H2_{(g)} (1 bar) | Pt_{(s)}}\]

Given: \[\ce{E^\circ(Sn^2+/sn) = -0.14 V, E^\circ H+/H2 = 0.00 V (log 10 = 1)}\]


Calculate e.m.f of the following cell at 298 K:

2Cr(s) + 3Fe2+ (0.1M) → 2Cr3+ (0.01M) + 3 Fe(s)

Given: E°(Cr3+ | Cr) = – 0.74 VE° (Fe2+ | Fe) = – 0.44 V


Calculate emf of the following cell at 25 °C :

Fe|Fe2+(0.001 M)| |H+(0.01 M)|H2(g) (1 bar)|Pt (s)

E°(Fe2+| Fe)0.44 V E°(H+ | H20.00 V


 
 

Calculate e.m.f. and ∆G for the following cell:

Mg (s) |Mg2+ (0.001M) || Cu2+ (0.0001M) | Cu (s)

`"Given :" E_((Mg^(2+)"/"Mg))^0=−2.37 V, E_((Cu^(2+)"/"Cu))^0=+0.34 V.`

 

 
 

In the representation of the galvanic cell, the ions in the same phase are separated by a _______.


Draw a neat and labelled diagram of the lead storage battery.


Standard hydrogen electrode operated under standard conditions of 1 atm Hpressure, 298 K, and pH = 0 has a cell potential of ____________.


The difference between the electrode potentials of two electrodes when no current is drawn through the cell is called ______.


Using the data given below find out the strongest reducing agent.

`"E"_("Cr"_2"O"_7^(2-)//"Cr"^(3+))^⊖` = 1.33 V  `"E"_("Cl"_2//"Cl"^-) = 1.36` V

`"E"_("MnO"_4^-//"Mn"^(2+))` = 1.51 V  `"E"_("Cr"^(3+)//"Cr")` = - 0.74 V


Consider the figure and answer the following question.

If cell ‘A’ has ECell = 0.5V and cell ‘B’ has ECell = 1.1V then what will be the reactions at anode and cathode?


Represent the cell in which the following reaction takes place.The value of E˚ for the cell is 1.260 V. What is the value of Ecell?

\[\ce{2Al (s) + 3Cd^{2+} (0.1M) -> 3Cd (s) + 2Al^{3+} (0.01M)}\]


Which is the correct order of second ionization potential of C, N, O and F in the following?


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The emf of a galvanic cell, with electrode potential of Zn2+ = - 0.76 V and that of Cu2+ = 0.34 V, is ______.


A voltaic cell is made by connecting two half cells represented by half equations below:

\[\ce{Sn^{2+}_{ (aq)} + 2e^- -> Sn_{(s)}}\], E0 = − 0.14 V

\[\ce{Fe^{3+}_{ (aq)} + e^- -> Fe^{2+}_{ (aq)}}\], E0 = + 0.77 V

Which statement is correct about this voltaic cell?


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