Advertisements
Advertisements
प्रश्न
Draw a neat and labelled diagram of the lead storage battery.
उत्तर
APPEARS IN
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 |
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: \[\ce{E^\circ(Sn^2+/sn) = -0.14 V, E^\circ H+/H2 = 0.00 V (log 10 = 1)}\]
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
Calculate e.m.f. and ∆G for the following cell:
Mg (s) |Mg2+ (0.001M) || Cu2+ (0.0001M) | Cu (s)
`"Given :" E_((Mg^(2+)"/"Mg))^0=−2.37 V, E_((Cu^(2+)"/"Cu))^0=+0.34 V.`
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.
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 :
Standard electrode potential is measured taking the concentrations of all the species involved in a half-cell is ____________.
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
The positive value of the standard electrode potential of Cu2+/Cu indicates that:
(i) this redox couple is a stronger reducing agent than the H+/H2 couple.
(ii) this redox couple is a stronger oxidising agent than H+/H2 .
(iii) Cu can displace H2 from acid.
(iv) Cu cannot displace H2 from acid.
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?
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?
Which is the correct order of second ionization potential of C, N, O and F in the following?
The potential of a hydrogen electrode at PH = 10 is
Standard reduction potentials (E°) of Cd2+, respectively which is the strongest reducing agent
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?
