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प्रश्न
For the given cell, \[\ce{Mg | Mg^{2+} || Cu^{2+} | Cu}\]
(i) \[\ce{Mg}\] is cathode
(ii) \[\ce{Cu}\] is cathode
(iii) The cell reaction is \[\ce{Mg^+ Cu^{2+} -> Mg^{2+} + Cu}\]
(iv) \[\ce{Cu}\] is the oxidising agent
उत्तर
(ii) \[\ce{Cu}\] is cathode
(iii) The cell reaction is \[\ce{Mg^+ Cu^{2+} -> Mg^{2+} + Cu}\]
Explanation:
(i) Left side of cell reaction represents oxidation half-cell i.e., oxidation of \[\ce{Mg}\] and right side of cell represents reduction half-cell reactions i.e., reduction of copper.
(ii) \[\ce{Cu}\] is reduced and reduction occurs at cathode.
(iii) \[\ce{Mg}\] is oxidized and oxidation occurs at anode.
(iv) Whole cell reaction can be written as
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संबंधित प्रश्न
Construct a labelled diagram for the following cell:
`Zn|Zn^(2+)(1M)||H^+(1M)|H_(2(g,1atm))|Pt`
In the electrochemical cell: Zn|ZnSO4 (0.01 M)||CuSO4 (1.0 M)|Cu, the emf of this Daniel cell is E1. When the concentration of ZnSO4 is changed to 1.0 M and that CuSO4 changed to 0.01 M, the emf changes to E2. From the above, which one is the relationship between E1 and E2?
Cell equation: \[\ce{A + 2B^- -> A^{2+} + 2B}\]
\[\ce{A^{2+} + 2e^- -> A}\] E0 = +0.34 V and log10 k = 15.6 at 300 K for cell reactions find E0 for \[\ce{B^+ + e^- -> B}\]
Use the data given in below find out the most stable ion in its reduced form.
`"E"_("Cr"_2"O"_7^(2-)//"Cr"^(3+))^⊖`= 1.33 V `"E"_("Cl"_2//"Cl"^-)^⊖` = 1.36 V
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Consider a cell given below:
\[\ce{Cu | Cu^{2+} || Cl^{-} | Cl_{2},Pt}\]
Write the reactions that occur at anode and cathode
Match the terms given in Column I with the items given in Column II.
| Column I | Column II |
| (i) Λm | (a) intensive property |
| (ii) ECell | (b) depends on number of ions/volume |
| (iii) K | (c) extensive property |
| (iv) ∆rGCell | (d) increases with dilution |
Match the items of Column I and Column II.
| Column I | Column II |
| (i) K | (a) I × t |
| (ii) Λm | (b) `Λ_m/Λ_m^0` |
| (iii) α | (c) `K/c` |
| (iv) Q | (d) `G^∗/R` |
Calculate the standard EMF ofa cell which involves the following cell reactions
\[\ce{Zn + 2 Ag+ -> Zn^{2+} + 2 Ag}\]
Given that \[\ce{E^{o}_{Zn/Zn^{2+}}}\] = 0.76 volt and \[\ce{E^{o}_{Ag/Ag^{+}}}\] = – 0.80 volt.
If the half-cell reaction A + e– → A– has a large negative reduction potential, it follow that:-
Read the passage given below and answer the questions that follow:
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Oxidation-reduction reactions are commonly known as redox reactions. They involve transfer of electrons from one species to another. In a spontaneous reaction, energy is released which can be used to do useful work. The reaction is split into two half-reactions. Two different containers are used and a wire is used to drive the electrons from one side to the other and a Voltaic/Galvanic cell is created. It is an electrochemical cell that uses spontaneous redox reactions to generate electricity. A salt bridge also connects to the half-cells. The reading of the voltmeter gives the cell voltage or cell potential or electromotive force. If \[\ce{E^0_{cell}}\] is positive the reaction is spontaneous and if it is negative the reaction is non-spontaneous and is referred to as electrolytic cell. Electrolysis refers to the decomposition of a substance by an electric current. One mole of electric charge when passed through a cell will discharge half a mole of a divalent metal ion such as Cu2+. This was first formulated by Faraday in the form of laws of electrolysis.
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- Is silver plate the anode or cathode? (1)
- What will happen if the salt bridge is removed? (1)
- When does electrochemical cell behaves like an electrolytic cell? (1)
- (i) What will happen to the concentration of Zn2+ and Ag+ when Ecell = 0. (1)
(ii) Why does conductivity of a solution decreases with dilution? (1)
OR
The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm2mol-1. Calculate the conductivity of this solution. (2)
