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प्रश्न
Using the data given below find out the strongest reducing agent.
`"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
पर्याय
Cl–
Cr
Cr3+
Mn2+
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उत्तर
Cr
Explanation:
A negative value of standard reduction potential for Cr3+ to Cr means that the redox couple is a stronger reducing agent.
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संबंधित प्रश्न
Draw a neat and well labelled diagram of primary reference electrode.
Arrange the following reducing agents in the order of increasing strength under standard state conditions. Justify the answer
|
Element |
Al(s) |
Cu(s) |
Cl(aq) |
Ni(s) |
|
Eo |
-1.66V |
0.34V |
1.36V |
-0.26V |
Calculate Ecell and ΔG for the following at 28°C :
Mg(s) + Sn2+( 0.04M ) → Mg2+( 0.06M ) + Sn(s)
E°cell = 2.23V. Is the reaction spontaneous ?
The standard e.m.f of the following cell is 0.463 V
`Cu|Cu_(1m)^(++)`
What is the standard potential of Cu electrode?
(A) 1.137 V
(B) 0.337 V
(C) 0.463 V
(D) - 0.463 V
Calculate e.m.f of the following cell at 298 K:
2Cr(s) + 3Fe2+ (0.1M) → 2Cr3+ (0.01M) + 3 Fe(s)
Given: E°(Cr3+ | Cr) = – 0.74 VE° (Fe2+ | Fe) = – 0.44 V
Given the standard electrode potentials,
K+/K = −2.93 V, Ag+/Ag = 0.80 V,
Hg2+/Hg = 0.79 V
Mg2+/Mg = −2.37 V, Cr3+/Cr = −0.74 V
Arrange these metals in their increasing order of reducing power.
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Standard hydrogen electrode operated under standard conditions of 1 atm H2 pressure, 298 K, and pH = 0 has a cell potential of ____________.
Which cell will measure standard electrode potential of copper electrode?
The difference between the electrode potentials of two electrodes when no current is drawn through the cell is called ______.
Use the data given in below find out which of the following is the strongest oxidising agent.
`"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
Consider the figure and answer the following question.
If cell ‘A’ has ECell = 0.5V and cell ‘B’ has ECell = 1.1V then what will be the reactions at anode and cathode?
Represent the cell in which the following reaction takes place.The value of E˚ for the cell is 1.260 V. What is the value of Ecell?
\[\ce{2Al (s) + 3Cd^{2+} (0.1M) -> 3Cd (s) + 2Al^{3+} (0.01M)}\]
The standard electrode potential of the two half cells are given below:
\[\ce{Ni^{2+} + 2e^{-} -> Ni, E_0 = - 0.25 Volt}\]
\[\ce{Zn^{2+} + 2e^{-} -> Zn, E_0 = - 0.77 Volt}\]
The voltage of cell formed by combining the two half cells would be?
A voltaic cell is made by connecting two half cells represented by half equations below:
\[\ce{Sn^{2+}_{ (aq)} + 2e^- -> Sn_{(s)}}\], E0 = − 0.14 V
\[\ce{Fe^{3+}_{ (aq)} + e^- -> Fe^{2+}_{ (aq)}}\], E0 = + 0.77 V
Which statement is correct about this voltaic cell?
