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प्रश्न
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 |
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उत्तर
| Column I | Column II |
| (i) F2 | (c) non metal which is the best oxidising agent |
| (ii) Li | (a) metal is the strongest reducing agent |
| (iii) Au3+ | (g) metal ion which is an oxidising agent |
| (iv) Br– | (e) anion that can be oxidised by Au3+ |
| (v) Au | (d) unreactive metal |
| (vi) Li+ | (b) metal ion which is the weakest oxidising agent |
| (vii) F– | (f) anion which is the weakest reducing agent |
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संबंधित प्रश्न
The conductivity of 0.20 M solution of KCl at 298 K is 0.025 S cm−1. Calculate its molar conductivity.
The conductivity of 0.001 mol L-1 solution of CH3COOH is 3.905× 10-5 S cm-1. Calculate its molar conductivity and degree of dissociation (α) Given λ°(H+)= 349.6 S cm2 mol-1 and λ°(CH3COO)= 40.9S cm2mol-1.
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| 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 `Lambda_m^0`.
Write mathematical expression of molar conductivity of the given solution at infinite dilution.
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)
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Choose the right option for your answer.
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\[\begin{array}{cc}
\end{array}\]\[\begin{bmatrix}
\ce{\Lambda^{\circ}_{H^+} = 350 S cm^2 mol^{-1}}\\
\ce{\Lambda^{\circ}_{CH_3COO^-} = 50 S cm^2 mol^{-1}}
\end{bmatrix}\]
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Conductivity of 2 × 10−3 M methanoic acid is 8 × 10−5 S cm−1. Calculate its molar conductivity and degree of dissociation if `∧_"m"^0` for methanoic acid, is 404 S cm2 mol−3.
