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प्रश्न
When acidulated water (dil.H2SO4 solution) is electrolysed, will the pH of the solution be affected? Justify your answer.
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उत्तर
Anode: \[\ce{2H2O(l) -> O2(g) + 4H + (aq) + 4e^{-}}\]
Cathode: \[\ce{4H^{+} + 4e^{-} -> 2H2}\]
Overall cell reaction: \[\ce{2H2O(l) -> O2(g) + 2H2(g)}\]
pH remains the same because concentration of H+ ions remains constant.
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संबंधित प्रश्न
Define limiting molar conductivity.
Why conductivity of an electrolyte solution decreases with the decrease in concentration ?
The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:
| Concentration/M | 0.001 | 0.010 | 0.020 | 0.050 | 0.100 |
| 102 × κ/S m−1 | 1.237 | 11.85 | 23.15 | 55.53 | 106.74 |
Calculate ∧m for all concentrations and draw a plot between ∧m and c1/2. Find the value of `∧_m^0`.
Calculate the degree of dissociation (α) of acetic acid if its molar conductivity (Λm) is 39.05 S cm2 mol−1.
(Given \[\ce{\lambda^{\circ}_{(H^+)}}\] = 349.6 S cm2 mol−1 and \[\ce{\lambda^{\circ}_{(CH_3COO^-)}}\] = 40.95 S cm2 mol−1)
Define the following terms :
Limiting molar conductivity
A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes FeSO4and ZnSO4 until 2.8g of Fe deposited at the cathode of cell X. How long did the current flow? Calculate the mass of Zn deposited at the cathode of cell Y.
(Molar mass: Fe=56g mol-1,Zn=65.3g mol-1,1F=96500C mol-1)
Molar conductivity denoted by the symbol Λm is related to the conductivity of the solution by the equation (k is the conductivity and c is the concentration).
Conductivity always decreases with decrease in concentration both, for weak and strong electrolytes because of the fact that ____________.
\[\ce{\Lambda^0_m(NH4OH)}\] is equal to ______.
\[\ce{Λ^0_m H2O}\] is equal to:
(i) \[\ce{Λ^0_m_{(HCl)} + \ce{Λ^0_m_{(NaOH)} - \ce{Λ^0_m_{(NaCl)}}}}\]
(ii) \[\ce{Λ^0_m_{(HNO_3)} + \ce{Λ^0_m_{(NaNO_3)} - \ce{Λ^0_m_{(NaOH)}}}}\]
(iii) \[\ce{Λ^0_{(HNO_3)} + \ce{Λ^0_m_{(NaOH)} - \ce{Λ^0_m_{(NaNO_3)}}}}\]
(iv) \[\ce{Λ^0_m_{(NH_4OH)} + \ce{Λ^0_m_{(HCl)} - \ce{Λ^0_m_{(NH_4Cl)}}}}\]
Why on dilution the m Λm of \[\ce{CH3COOH}\] increases very fast, while that of \[\ce{CH3COONa}\] increases gradually?
Match the items of Column I and Column II on the basis of data given below:
`E_("F"_2//"F"^-)^Θ` = 2.87 V, `"E"_(("Li"^(+))//("Li"^-))^Θ` = − 3.5V, `"E"_(("Au"^(3+))//("Au"))^Θ` = 1.4 V, `"E"_(("Br"_(2))//("Br"^-))^Θ` = 1.09 V
| Column I | Column II |
| (i) F2 | (a) metal is the strongest reducing agent |
| (ii) Li | (b) metal ion which is the weakest oxidising agent |
| (iii) Au3+ | (c) non metal which is the best oxidising agent |
| (iv) Br– | (d) unreactive metal |
| (v) Au | (e) anion that can be oxidised by Au3+ |
| (vi) Li+ | (f) anion which is the weakest reducing agent |
| (vii) F– | (g) metal ion which is an oxidising agent |
Assertion: Copper sulphate can be stored in zinc vessel.
Reason: Zinc is less reactive than copper.
An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to :-
The molar conductivity of CH3COOH at infinite dilution is 390 Scm2/mol. Using the graph and given information, the molar conductivity of CH3COOK will be:
Given below are two statements:
Statements I: The limiting molar conductivity of KCl (strong electrolyte) is higher compared to that of CH3COOH (weak electrolyte).
Statement II: Molar conductivity decreases with decrease in concentration of electrolyte.
In the light of the above statements, choose the most appropriate answer from the options given below:
Which of the following solutions of KCl will have the highest value of molar conductivity?
The solution of two electrolytes A and B are diluted. ^m of B increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte? Give a reason.
