Calculate emf of the following cell at 25^{\circ} {\C} : ...

CBSE Class 12 Chemistry 2015 Solved Paper

Question : 21 of 26

Marks: +1, -0

Calculate emf of the following cell at

25^{∘} {\C}

\; {\Fe}{| {\Fe}^{2+}(0.001 {\M}) \∥ {\H}^{+}(0.01 {\M}) |}

{\H}_2( {\g})(\bo {1} \;\text " bar "\;) ∣ {\Pt}( {\ s})

\; {\E}^{∘} ( {\Fe}^{2+} ∣ {\Fe}) =-0.44 {\V} {\E}^{∘}

( {\H}^{+} ∣ {\H}_2) =0.00 {\V}

Solution:

The cell reaction:

\; {\Fe}( {\ s})+2 {\H}^{+}( {\aq}) \ {  ────▸} {\Fe}^{2+}( {\aq})

+ {\H}_2( {\g})

\; {\E}°_{\;\text "cell "\;} = {\E}_{ {\c}}^{∘}- {\E}_{ {\a}}^{∘}

\;\; =[0-(-0.44)] {\V}=0.44 {\V}

\; {\E}_{\;\text "cell "\;}= {\E°}_{\;\text "cell "\;}-\;{0.059}/{2} \log \;{ [ {\Fe}^{2+}] }/{ [ {\H}^{+}] ^2}

\; {\E}_{\;\text "cell "\;}=0.44 {\V}-\;{0.059}/{2} \log \;{(0.001)}/{(0.01)^2}

\;\; =0.44 {\V}-\;{0.059}/{2} \log (10)

\;=0.44 {\V}-0.0295 {\V}

\;\; ≈ 0.410 {\V}

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