Calculate e.m.f. of the following cell at 298 {\K} : 2 {\Cr}( { s})+3 {\Fe}^{2+}(0.1 {\M}) \to 2 {\Cr}^{3+}(0.01 {\M}) +3 {\Fe}( { s}) \text{ Given : } {\E}^\circ_{ ( {\Cr}^{3+} ∣ {\Cr}) }=-0.74 {\V} {\E}^{\circ} ( {\Fe}^{2+} ∣ {\Fe}) =-0.44 {\V} 3

CBSE Class 12 Chemistry 2016 Delhi Set 1 Solved Paper

Question : 16 of 26

Marks: +1, -0

Calculate e.m.f. of the following cell at

298 {\K}

\;2 {\Cr}( {\ s})+3 {\Fe}^{2+}(0.1 {\M}) → 2 {\Cr}^{3+}(0.01 {\M})

+3 {\Fe}( {\ s})

\;\;\text " Given : "\; {\E}°_{ ( {\Cr}^{3+} ∣ {\Cr}) }=-0.74 {\V} {\E}^{∘}

( {\Fe}^{2+} ∣ {\Fe}) =-0.44 {\V} 3

Solution:

{\E}°_{\;\text "celi "\;} \;= {\E}_c^0- {\E}_a^0

\;=(-0.44)-(-0.74) {\V}

\;=0.30 {\V}

{\E}_{\;\text "cell "\;}\;= {\E}°_{\;\text "cell "\;} -\;{0.059}/{n} \log \;{ [ {\Cr}^{3+}] ^2}/{ [ {\Fe}^{2+}] ^3}

{\E}_{\;\text "cell "\;}\;= {\E}°_{\;\text "cell "\;}-\;{0.059}/{6} \log \;{[0.01]^2}/{[0.1]^3}

\;=0.30- (\;{-0.059}/{6})

\;=0.3098 {\V}

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