Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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Calculate the Emf of the Following Cell at 298 K - Chemistry

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Calculate the emf of the following cell at 298 K:

`Fe(s)|Fe^(2+)(0.001M)||H^+(1M)|H_2 "(g)(1bar) ",Pt(s)`

`("Given " E_(cell)^@=+0.44V)`

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Solution

At anode: Fe → Fe2+ + 2e-

At cathode: 2H+ + 2e- → H2

So, total number of electrons (n) transferred = 2

Given: Eocell = +0.44 Volt

Temperature (T) = 298 K

`E_(cell)=E_(cell)^@-((2.303RT)/(nF))log a_(oxi)/a_(red)`

`E_(cell)=E_(cell)^@-((0.05916V)/(n))log a_(oxi)/a_(red)=>E_(cell)=0.44-0.0591V/2 log 0.001/1`

Therefore, Ecell = 0.44(−0.02955 × − 3) = 0.44 + 0.08865 = 0.53 Volt.

Concept: Galvanic Cells - Introduction
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