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प्रश्न
Given the standard electrode potentials,
\[\ce{K+/K}\] = −2.93 V, \[\ce{Ag+/Ag}\] = 0.80 V,
\[\ce{Hg^{2+}/Hg}\] = 0.79 V
\[\ce{Mg^{2+}/Mg}\] = −2.37 V, \[\ce{Cr^{3+}/Cr}\] = −0.74 V
Arrange these metals in their increasing order of reducing power.
उत्तर
The reducing power of a metal depends on its oxidation potential. The higher the oxidation potential, the greater its tendency to be oxidized and, hence, the greater its reducing power. Hence the order of increasing of reducing the power of the given metals will be as follows:
\[\ce{Ag < Hg < Cr < Mg < K}\]
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संबंधित प्रश्न
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 emf of the following cell at 25°C:
\[\ce{Sn/Sn^2+ (0.001 M) || H+ (0.01 M) | H2_{(g)} (1 bar) | Pt_{(s)}}\]
Given: \[\ce{E^\circ(Sn^2+/sn) = -0.14 V, E^\circ H+/H2 = 0.00 V (log 10 = 1)}\]
Calculate emf of the following cell at 25 °C :
Fe|Fe2+(0.001 M)| |H+(0.01 M)|H2(g) (1 bar)|Pt (s)
E°(Fe2+| Fe)= −0.44 V E°(H+ | H2) = 0.00 V
Galvanic or a voltaic cell converts the chemical energy liberated during a redox reaction to ____________.
Standard electrode potential is measured taking the concentrations of all the species involved in a half-cell is ____________.
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 ______.
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
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
What does the negative sign in the expression `"E"^Θ ("Zn"^(2+))//("Zn")` = − 0.76 V mean?
Assertion: Cu is less reactive than hydrogen.
Reason: `E_((Cu^(2+))/(Cu))^Θ` is negative.
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 potential of a hydrogen electrode at PH = 10 is
The emf of a galvanic cell, with electrode potential of Zn2+ = - 0.76 V and that of Cu2+ = 0.34 V, is ______.
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?
