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Reduction potential of two metals M1 and M2 are EXM12+|M10 = −2.3 V and EXM22+|M20 = 0.2 V. Predict which one is better for coating the surface of iron.Given: EXFe2+|Fe0 = −0.44 V - Chemistry

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प्रश्न

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

एका वाक्यात उत्तर

उत्तर

oxidation potential of M1 is more +ve than the oxidation potential of Fe which indicates that it will prevent iron from rusting.

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पाठ 9: Electro Chemistry - Evaluation [पृष्ठ ६६]

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सामाचीर कलवी Chemistry - Volume 1 and 2 [English] Class 12 TN Board
पाठ 9 Electro Chemistry
Evaluation | Q 17. | पृष्ठ ६६

संबंधित प्रश्‍न

Can you store copper sulphate solutions in a zinc pot?


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` = 1.1V for Daniel cell. Which of the following expressions are correct description of state of equilibrium in this cell?

(i) 1.1 = `K_c`

(ii) `(2.303RT)/(2F) logK_c` = 1.1

(iii) `log K_c = 2.2/0.059`

(iv) `log K_c` = 1.1


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


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

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


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Ag + I → Agl + e; E° = – 0.152 V

Ag → Ag+ + e; E° = – 0.800 V

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\[\ce{Ag+ + e- -> Ag}\], `"E"_("Ag"^+//"Ag")^circ` = - 0.3995 V

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