Advertisements
Advertisements
Question
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?
Advertisements
Solution
In the electrolysis process of sodium chloride, oxidation of water at anode requires more potential. So \[\ce{Cl-}\] is oxidized anode instead of water.
APPEARS IN
RELATED QUESTIONS
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.
Calculate E°cell for the following reaction at 298 K:
2Al(s) + 3Cu+2(0.01M) → 2Al+3(0.01M) + 3Cu(s)
Given: Ecell = 1.98V
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 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
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.
Standard electrode potential is measured taking the concentrations of all the species involved in a half-cell is ____________.
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
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 ______.
