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Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because ____________.

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प्रश्न

Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because ____________.

पर्याय

  • Zinc is lighter than iron

  • Zinc has lower melting point than iron

  • Zinc has lower negative electrode potential than iron

  • Zinc has higher negative electrode potential than iron

MCQ
रिकाम्या जागा भरा
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उत्तर

Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because Zinc has higher negative electrode potential than iron.

Explanation:

`"E"_("Zn"^(2+)|"Zn")^0` = – 0.76 V and `"E"_("Fe"^(2+)|"Fe")^0` = 0.44 V

Zinc has higher negative electrode potential than iron, iron cannot be coated on zinc.

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पाठ 9: Electro Chemistry - Evaluation [पृष्ठ ६४]

APPEARS IN

सामाचीर कलवी Chemistry - Volume 1 and 2 [English] Class 12 TN Board
पाठ 9 Electro Chemistry
Evaluation | Q 14. | पृष्ठ ६४

संबंधित प्रश्‍न

Derive a relation between ΔH and ΔU for a chemical reaction. Draw neat labelled diagram of calomel electrode. Resistance and conductivity of a cell containing 0.001 M KCI solution at 298K are 1500Ω and 1.46x10-4 S.cm-1 respectively.


Can you store copper sulphate solutions in a zinc pot?


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


Describe the construction of Daniel cell. Write the cell reaction.


A copper electrode is dipped in 0.1 M copper sulphate solution at 25°C. Calculate the electrode potential of copper.
[Given: \[\ce{E^0_{{Cu^{2+}|Cu}}}\] = 0.34 V]


Electrode potential for Mg electrode varies according to the equation

`E_(Mg^(2+)  |  Mg) = E_(Mg^(2+)  |  Mg)^Θ - 0.059/2 log  1/([Mg^(2+)])`. The graph of `E_(Mg^(2+)  |  Mg)` vs `log [Mg^(2+)]` is ______.


Which of the following statement is correct?


`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.


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

Assertion: ECell should have a positive value for the cell to function.

Reason: `"E"_("cathode") < "E"_("anode")`


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