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Derive an expression for the Nernst equation.

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प्रश्न

Derive an expression for the Nernst equation.

संक्षेप में उत्तर
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उत्तर

Nernst equation is the one which relates the cell potential and the concentration of the species involved in an electrochemical reaction.

Let us consider an electrochemical cell for which the overall redox reaction is,

\[\ce{{x}A + {y}B ⇌ {l}C + {m}D}\]

The reaction quotient Q is,

Q = `(["C"]^"l" ["D"]^"m")/(["A"]^"x" ["B"]^"y")` .......(1)

We know that,

∆G = ∆G° + RT ln Q .......(2)

∆G = – nFEcell

∆G° = `-"nFE"_"cell"^∘`

equation (1) becomes

– nFEcell = `-"nFE"_"cell"^∘ + "RT" ln "Q"`

Subsitute the Q value in equation (2)

`-"nFE"_"cell" = -"nFE"_"cell"^∘ + "RT" ln  (["C"]^"l" ["D"]^"m")/(["A"]^"x" ["B"]^"y")` …..(3)

Divide the whole equation (3) by (– nF)

Ecell = `"E"_"cell"^∘ - "RT"/"nF" ln  ((["C"]^"l" ["D"]^"m")/(["A"]^"x" ["B"]^"y"))`

Ecell = `"E"_"cell"^∘ - (2.303  "RT")/"nF" log  ((["C"]^"l" ["D"]^"m")/(["A"]^"x" ["B"]^"y"))` .........(4)

This is called the Nernst equation.

At 25°C (298 K) equation (4) becomes,

Ecell = `"E"_"cell"^∘ - (2.303 xx 8.314 xx 298)/("n"(96500)) log  ((["C"]^"l" ["D"]^"m")/(["A"]^"x" ["B"]^"y"))`

Ecell = `"E"_"cell"^∘ - 0.0591/"n" log  ((["C"]^"l" ["D"]^"m")/(["A"]^"x" ["B"]^"y"))`

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Thermodynamics of Cell Reactions
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अध्याय 9: Electro Chemistry - Evaluation [पृष्ठ ६७]

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सामाचीर कलवी Chemistry - Volume 1 and 2 [English] Class 12 TN Board
अध्याय 9 Electro Chemistry
Evaluation | Q 24. | पृष्ठ ६७

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