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प्रश्न
In the representation of the galvanic cell, the ions in the same phase are separated by a _______.
पर्याय
single vertical line
comma
double vertical line
semicolon
उत्तर
In the representation of the galvanic cell, the ions in the same phase are separated by a comma.
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संबंधित प्रश्न
Draw a neat and well labelled diagram of primary reference electrode.
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
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.
Draw a neat and labelled diagram of the lead storage battery.
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 ____________.
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
What does the negative sign in the expression `"E"^Θ ("Zn"^(2+))//("Zn")` = − 0.76 V mean?
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?
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 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:
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?
