Advertisements
Advertisements
प्रश्न
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?
पर्याय
E1 < E2
E1 > E2
E2 ≥ E1
E1 = E2
उत्तर
E1 > E2
Explanation:
`"E"_"cell" = "E"_"cell"^0 - 0.0591/2 log (["Zn"^(2+)])/(["Cu"^(2+)])`
E1 = `"E"_"cell"^0 - 0.0591/2 log 10^-2/1`
E1 = `"E"_"cell"^0 + 0.0591` .....................(1)
\[\ce{Zn_{(s)} -> Zn^{2+}_{( aq)} + 2e^-}\]
\[\ce{Cu^{2+}_{( aq)} + 2e^- -> Cu_{(s)}}\]
\[\ce{Zn_{(s)} + Cu^{2+}_{( aq)} -> Zn^{2+}_{( aq)} + Cu_{(s)}}\]
E2 = `"E"_"cell"^0 - 0.0591/2 log 1/10^-2`
E2 = `"E"_"cell"^0 - 0.0591` .....................(2)
∴ E1 > E2
APPEARS IN
संबंधित प्रश्न
A solution of CuSO4 is electrolysed using a current of 1.5 amperes for 10 minutes. What mass of Cu is deposited at cathode? [Atomic mass of Cu = 63.7]
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 current strength of 3.86 A was passed through molten Calcium oxide for 41minutes and 40 seconds. The mass of Calcium in grams deposited at the cathode is (atomic mass of Ca is 40g/mol and 1F = 96500 C).
Electrode potential for Mg electrode varies according to the equation
`E_(Mg^(2+) | Mg) = E_(Mg^(2+) | Mg)^Θ - 0.059/2 log 1/([Mg^(2+)])`. The graph of `E_(Mg^(2+) | Mg)` vs `log [Mg^(2+)]` is ______.
`E_(cell)^Θ` for some half cell reactions are given below. On the basis of these mark the correct answer.
(a) \[\ce{H^{+} (aq) + e^{-} -> 1/2 H_2 (g); E^Θ_{cell} = 0.00V}\]
(b) \[\ce{2H2O (1) -> O2 (g) + 4H^{+} (aq) + 4e^{-}; E^Θ_{cell} = 1.23V}\]
(c) \[\ce{2SO^{2-}_{4} (aq) -> S2O^{2-}_{8} (aq) + 2e^{-}; E^Θ_{cell} = 1.96V}\]
(i) In dilute sulphuric acid solution, hydrogen will be reduced at cathode.
(ii) In concentrated sulphuric acid solution, water will be oxidised at anode.
(iii) In dilute sulphuric acid solution, water will be oxidised at anode.
(iv) In dilute sulphuric acid solution, \[\ce{SO4^{2-}}\] ion will be oxidised to tetrathionate ion at anode.
Consider the following diagram in which an electrochemical cell is coupled to an electrolytic cell. What will be the polarity of electrodes ‘A’ and ‘B’ in the electrolytic cell?
The correct order of the mobility of the alkali metal ions. In aqueous solultion is
Read the passage given below and answer the questions that follow:
|
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.
|
- 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)
If the value of Ksp for Hg2Cl2 (s) is X then the value of X will be ____ where pX = - log X.
Given:
\[\ce{Hg2Cl2 + 2e- -> 2Hg(l) + 2Cl-}\], E° = 0.27 V
\[\ce{Hg+2 + 2e- -> 2Hg(l)}\] E° = 0.81 V
Calculate the λ0m for Cl- ion from the data given below:
∧0m MgCl2 = 258.6 Scm2 mol-1 and λ0m Mg2+ = 106 Scm2 mol-1
