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प्रश्न
Consider the change in the oxidation state of Bromine corresponding to different emf values as shown in the expression below:
\[\ce{BrO^-_4 ->[1.82 V] BrO^-_3 ->[1.5 V] HBrO ->[1.595 V] Br2 ->[1.0652 V] Br^-}\]
Then the species undergoing disproportionation is:
विकल्प
Br2
\[\ce{BrO^-_4}\]
\[\ce{BrO^-_3}\]
HBrO
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उत्तर
HBrO
Explanation:
(Ecell)A = – 1.82 + 1.5 = – 0.32 V
(Ecell)B = – 1.5 + 1.595 = + 0.095 V
(Ecell)C = 1.595 + 1.0652 = – 0.529 V
∴ The species undergoing disproportionation is HBrO.
संबंधित प्रश्न
Arrange the following metals in the order in which they displace each other from the solution of their salts.
\[\ce{Al, Cu, Fe, Mg}\] and \[\ce{Zn}\]
For the electrochemical cell:
\[\ce{M | M+ || X- | X}\];
\[\ce{E^{\circ}_{{M^{+}/{M}}}}\] = 0.44 V,
\[\ce{E^{\circ}_{X/X^-}}\] = 0.33 V
Which of the following is TRUE for this data?
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]
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 \[\ce{B^+ + e^- -> B}\]
Two metals M1 and M2 have reduction potential values of −xV and +yV respectively. Which will liberate H2 and H2SO4.
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
`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.
A galvanic cell has electrical potential of 1.1V. If an opposing potential of 1.1V is applied to this cell, what will happen to the cell reaction and current flowing through the cell?
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 |
What should be the signs (positive/negative) for \[\ce{E^0_{cell}}\] and ΔG0 for a spontaneous redox reaction occurring under standard conditions?
