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प्रश्न
What is electrode potential?
What do you understand by electrode potential?
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उत्तर
The tendency of an electrode to lose or gain electrons when it is in contact with its own ions in solution is called electrode potential. The electrode potentials are responsible for the e.m.f. of a galvanic cell.
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संबंधित प्रश्न
Derive a relation between ΔH and ΔU for a chemical reaction. Draw neat labelled diagram of calomel electrode. Resistance and conductivity of a cell containing 0.001 M KCI solution at 298K are 1500Ω and 1.46x10-4 S.cm-1 respectively.
How many moles of electrons are required for reduction of 2 moles of Zn2+ to Zn?
A solution of Cu(NO3)2 is electrolyzed between platinum electrodes using 0.1 Faraday electricity. How many moles of Cu will be deposited at the cathode?
A certain current liberated 0.504 gm of hydrogen in 2 hours. How many grams of copper can be liberated by the same current flowing for the same time through copper sulphate solution.
Define cathode
Describe the electrolysis of molten NaCl using inert electrodes.
Which of the following statement is correct?
An electrochemical cell can behave like an electrolytic cell when ______.
Use the data given in below find out which option the order of reducing power is correct.
`"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
Use the data given in below find out the most stable oxidised species.
`E^0 (Cr_2O_1^(2-))/(Cr_(3+))` = 1.33 V `E^0 (Cl_2)/(Cl^-)` = 1.36 V
`E^0 (MnO_4^-)/(MN^(2+))` = 1.51 V `E^0 (Cr^(3+))/(Cr)` = – 0.74 V
Depict the galvanic cell in which the cell reaction is \[\ce{Cu + 2Ag^+ -> 2Ag + Cu^{2+}}\]
How will the pH of brine (aq. \[\ce{NaCl}\] solution) be affected when it is electrolysed?
Match the terms given in Column I with the units given in Column II.
| Column I | Column II |
| (i) Λm | (a) S cm-¹ |
| (ii) ECell | (b) m-¹ |
| (iii) K | (c) S cm2 mol-¹ |
| (iv) G* | (d) V |
Standard reduction electrode potentials of three metals A, B and C are respectively + 0.5 V, – 3.0 V. and – 1.2 V. The reducing powers of there metal is.
Cell reaction is spontaneous when
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)
The number of moles of electrons passed when the current of 2 A is passed through a solution of electrolyte for 20 minutes is ______.
State the term for the following:
Two metal plates or wires through which the current enters and leaves the electrolytic cell.
