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प्रश्न
What happens if external potential applied becomes greater than E°cell of electrochemical cell?
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उत्तर
If the external potential applied becomes greater than the E°cell of electrochemical cell, then the reaction gets reversed and the electrochemical cell acts as an electrolytic cell and vice-versa.
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]
At 25°C, the emf of the following electrochemical cell.
\[\ce{Ag_{(s)} | Ag^+ (0.01 M) | | Zn^{2+} {(0.1 M)} | Zn_{(s)}}\] will be:
(Given \[\ce{E^0_{cell}}\] = −1.562 V)
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).
A gas X at 1 atm is bubbled through a solution containing a mixture of 1 MY− and 1 MZ− at 25°C. If the reduction potential of Z > Y > X, then ______.
Define cathode
`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.
Can absolute electrode potential of an electrode be measured?
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 |
Assertion: ECell should have a positive value for the cell to function.
Reason: `"E"_("cathode") < "E"_("anode")`
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
