Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
A metal’s oxidation potential determines its reducing power. Its tendency to be oxidised and, thus, its reducing power increase with its oxidation potential. Therefore, the following will be the order in which the power of the specified metals increases or decreases:
Ag < Hg < Cr < Mg < K
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 ?
Can copper sulphate solution be stored in an iron vessel? Explain.
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 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: \[\ce{E^\circ(Sn^2+/sn) = -0.14 V, E^\circ H+/H2 = 0.00 V (log 10 = 1)}\]
Depict the galvanic cell in which the reaction \[\ce{Zn(s) + 2Ag+(aq) → Zn^{2+}(aq) + 2Ag(s)}\] takes place. Further show:
- Which of the electrode is negatively charged?
- The carriers of the current in the cell.
- Individual reaction at each electrode.
Calculate the emf of the following cell at 25°C :
Galvanic or a voltaic cell converts the chemical energy liberated during a redox reaction to ____________.
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
Value of standard electrode potential for the oxidation of \[\ce{Cl-}\] ions is more positive than that of water, even then in the electrolysis of aqueous sodium chloride, why is \[\ce{Cl-}\] oxidised at anode instead of water?
Which reference electrode is used to measure the electrode potential of other electrodes?
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?
Standard electrode potential of three metals X, Y and Z are –1.2 V, +0.5 V and –3.0 V, respectively. The reducing power of these metals will be ______.
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?
