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 → Fe^{2+} + 2e^{-}

**At cathode:** 2H^{+} + 2e^{-} → H_{2}

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

Given: E^{o}_{cell} = +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, E_{cell} = 0.44(−0.02955 × − 3) = 0.44 + 0.08865 = 0.53 Volt.

Concept: Galvanic Cells - Introduction

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