हिंदी

Calculate Ecell∘ of the following galvanic cell: Mg(s) / Mg2+(1 M) // Ag+ (1 M) / Ag(s) if EMg∘ = – 2.37 V and EAg∘ = 0.8 V.

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प्रश्न

Calculate `"E"_"cell"^circ` of the following galvanic cell:

Mg(s) / Mg2+(1 M) // Ag+ (1 M) / Ag(s) if `"E"_"Mg"^circ` = – 2.37 V and `"E"_"Ag"^circ` = 0.8 V. Write cell reactions involved in the above cell. Also mention if cell reaction is spontaneous or not.

संख्यात्मक
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उत्तर

Given: `"E"_"Mg"^circ` = – 2.37 V and `"E"_"Ag"^circ` = 0.8 V

To find: Standard cell potential

Formula: `"E"_"cell"^circ = "E"_"cathode"^circ - "E"_"anode"^circ`

Calculation: `"E"_"cell"^circ = "E"_"cathode"^circ - "E"_"anode"^circ`

`"E"_"cell"^circ = "E"_"Ag"^circ - "E"_"Mg"^circ`

= (0.8 V) - (- 2.37 V) = 3.17 V

The standard cell potential for the reaction is 3.17 V.

Cell reactions:

Electrode reactions are

At anode: \[\ce{Mg_{(s)} -> Mg^{2+}_{ (aq)} + 2e-}\]
At cathode: \[\ce{Ag^+_{ (aq)} + e^- -> Ag_{(s)}}\]
Overall cell reaction- \[\ce{Mg_{(s)} + 2Ag^+_{ (aq)} -> Mg^{2+}_{ (aq)} + 2Ag_{(s)}}\]

Since the standard cell potential is positive, the cell reaction is spontaneous.

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अध्याय 5: Electrochemistry - Long answer questions

संबंधित प्रश्न

Consider the half reactions with standard potentials.

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  3. \[\ce{Pb^{2\oplus}_{(aq)} + 2e^{\ominus} -> Pb_{(s)} E^\circ = -0.13 V}\]
  4. \[\ce{Fe^{2\oplus} + 2e^{\ominus} -> Fe_{(s)} E^\circ = -0.44 V}\]

The strongest oxidising  and reducing agents respectively are ______.


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What is standard cell potential for the reaction

\[\ce{3Ni_{(s)} + 2Al^{3+} (1M) → 3Ni^{2+} (1M) + 2Al(s)}\], if `E_"Ni"^circ` = –0.25 V and  `"E"_("Al")^circ` = –1.66 V?


Answer the following in one or two sentences.

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Answer the following:

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Given: `"E"_"Sn"^circ` = - 0.136, `"E"_"Ag"^circ` = 0.800 V


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Given: `"E"_("Cr"^(3+)//"Cr")^0` = −0.74 V,

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Element `"E"^0 ("V")`
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Hg +0.79
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