CBSE Class 12 Chemistry 2013 Solved Paper

Question : 21 of 30

Marks: +1, -0

Calculate the emf of the following cell at

298 {\K}

\; {\Fe}( {\ s}){| {\Fe}^{2+}(0.001 {\M}) || {\H}^{+}(1 {\M}) |} {\H}^2( {\g})(1 {\bar}),

\; {\Pt}( {\ s})

\; (\;\text " Given "\; {{\E}^{°}}_{\;\text "cell "\;}=+0.44 {\V})

Solution:

According to given equation:

{Fe}({s})+2 {H}^{+}({aq}) \→{Fe}^{+2}({aq})+{H}_2({g})

{E}°_{\text "cell"} =0.44 { V}

By applying Nernst equation:

{E}_{\text "cell "}={E}°_{\text "cell "} -{0.0591}/{n} \log {{Fe}^{+2}}/{[{H}^{+}]^2}

E_{\text "cell "}=0.44-{0.0591}/{2} \log {0.001}/{(1)^2}

{E}_{\text "cell "}=0.44-0.0295 \log 10^{-3}

E_{\text "cell "}=0.44-0.0295(-3 \log 10)

where

[\log 10=1]

=0.44-0.0295(-3 \× 1)

=0.44+0.0885

{E}_{\text "cell "}=0.528 {V}

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