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प्रश्न
|
`E_((M^(2+)/M)` |
Cr | Mn | Fe | Co | Ni | Cu |
| -0.91 | -1.18 | -0.44 | -0.28 | -0.25 | -0.34 |
From the given data of E0 values, answer the following questions :
(1) Why is `E_(((Cu^(2+))/(Cu)))` value exceptionally positive
(2) Why is `E_(((Mn^(2+))/(Mn)))` value is highly negative as compared to other elements
(3) Which is the stronger reducing agents Cr2+ or Fe2+ ? Give Reason.
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उत्तर
(i) The `E_((M^(2+)/M)` value of a metal depends on the energy changes involved in the formation of the M2+ion:
1. Sublimation:- The energy required for converting one mole of an atom from the solid state to the gaseous state
M(s) → M(g) ∆s H (Sublimation energy)
2. Ionisation:- The energy required to take out electrons from one mole of atoms in the gaseous state to form the corresponding cation in the gaseous state
M(g) → M2+(g) ∆iH (Ionization energy)
3. Hydration: The energy released when one mole of ions are hydrated
M2+(g) → M2+(aq) ∆hydH (Hydration energy)
Now, copper has high energy of atomisation and low hydration energy. Hence, the `E_(((Cu^(2+))/(Cu)))` is exceptionally positive.
(ii) The `E_(((Mn^(2+))/(Mn)))` value is highly negative as compared to other elements because of the extra stability of Mn2+ ion due to half-filled d-orbitals.
(iii) The following reactions are involved when Cr2+ and Fe2+ act as reducing agents:
Cr2+ → Cr3+ + e−
Fe2+ → Fe3+ + e−
The `E_(Cr^(3+))` value is −0.41 V and `E_(Fe^(3+))` is +0.77 V. This means that Cr2+ can be easily oxidised to Cr3+, but Fe2+ does not get oxidised to Fe3+ easily. Therefore, Cr2+ is a stronger reducing agent than Fe3+.
संबंधित प्रश्न
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| Column I (Catalyst) | Column II (Process) |
| (i) \[\ce{Ni}\] in the presence of hydrogen | (a) Zieglar Natta catalyst |
| (ii) \[\ce{Cu2C12}\] | (b) Contact process |
| (iii) \[\ce{V2O5}\] | (c) Vegetable oil to ghee |
| (iv) Finely divided iron | (d) Sandmeyer reaction |
| (v) \[\ce{TiCl4 + Al (CH3)3}\] | (e) Haber's Process |
| (f) Decomposition of KCIO3 |
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