Advertisements
Advertisements
Question
The cell constant of a conductivity cell is 0.146 cm-1. What is the conductivity of 0.01 M solution of an electrolyte at 298 K, if the resistance of the cell is 1000 ohm?
Advertisements
Solution
cell constant G* = k × R
k = `("G""*")/"R" = 0.146/ 1000 = 1.46 xx 10-4 "Scm"^-1.`
APPEARS IN
RELATED QUESTIONS
Why cannot we store AgNO3 solution in copper vessel?
E°cell for the given redox reaction is 2.71 V
Mg(s) + Cu2+ (0.01 M) → Mg2+ (0.001 M) + Cu(s)
Calculate Ecell for the reaction. Write the direction of flow of current when an external opposite potential applied is
(i) less than 2.71 V and
(ii) greater than 2.71 V
For the electrochemical cell:
\[\ce{M | M+ || X- | X}\];
\[\ce{E^{\circ}_{{M^{+}/{M}}}}\] = 0.44 V,
\[\ce{E^{\circ}_{X/X^-}}\] = 0.33 V
Which of the following is TRUE for this data?
Reduction potential of two metals M1 and M2 are \[\ce{E^0_{{M_1^{2+}|M_1}}}\] = −2.3 V and \[\ce{E^0_{{M_2^{2+}|M_2}}}\] = 0.2 V. Predict which one is better for coating the surface of iron.
Given: \[\ce{E^0_{{Fe^{2+}|Fe}}}\] = −0.44 V
Use the data given in below find out the most stable ion in its reduced form.
`"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
`E_(cell)^Θ` for some half cell reactions are given below. On the basis of these mark the correct answer.
(a) \[\ce{H^{+} (aq) + e^{-} -> 1/2 H_2 (g); E^Θ_{cell} = 0.00V}\]
(b) \[\ce{2H2O (1) -> O2 (g) + 4H^{+} (aq) + 4e^{-}; E^Θ_{cell} = 1.23V}\]
(c) \[\ce{2SO^{2-}_{4} (aq) -> S2O^{2-}_{8} (aq) + 2e^{-}; E^Θ_{cell} = 1.96V}\]
(i) In dilute sulphuric acid solution, hydrogen will be reduced at cathode.
(ii) In concentrated sulphuric acid solution, water will be oxidised at anode.
(iii) In dilute sulphuric acid solution, water will be oxidised at anode.
(iv) In dilute sulphuric acid solution, \[\ce{SO4^{2-}}\] ion will be oxidised to tetrathionate ion at anode.
For the given cell, \[\ce{Mg | Mg^{2+} || Cu^{2+} | Cu}\]
(i) \[\ce{Mg}\] is cathode
(ii) \[\ce{Cu}\] is cathode
(iii) The cell reaction is \[\ce{Mg^+ Cu^{2+} -> Mg^{2+} + Cu}\]
(iv) \[\ce{Cu}\] is the oxidising agent
Depict the galvanic cell in which the cell reaction is \[\ce{Cu + 2Ag^+ -> 2Ag + Cu^{2+}}\]
Match the terms given in Column I with the items given in Column II.
| Column I | Column II |
| (i) Λm | (a) intensive property |
| (ii) ECell | (b) depends on number of ions/volume |
| (iii) K | (c) extensive property |
| (iv) ∆rGCell | (d) increases with dilution |
If the half-cell reaction A + e– → A– has a large negative reduction potential, it follow that:-
