CBSE Class 12 Chemistry 2022 Term 2 Delhi Set 3

Question : 5 of 12

Marks: +1, -0

(a) Calculate

∆_r {\G}^{∘}

and

\log {\K}_c

for the following cell:

{\Ni}(s)+2 {\Ag}^{+}(a q) → {\Ni}^{2+}(a q)+2 {\Ag}( {\ s})

Given that

{E^{∘}}_\cell

=1.05 {\V}, 1 {\F}=96,500 {\C} {\mol}^{-1}

OR
(b) Calculate the e.m.f. of the following cell at

298 {\K}

{\Fe}(s){| {\Fe}^{2+}(0.001 {\M}) |}{| {\H}^{+}(0.01 {\M}) |} {\H}_2(g)

(1 bar) | Pt (s)
Given that

{ {\E}^{∘}}_\cell

=0.44 {\V}

[\log 2=0.3010, \log 3=0.4771, \log 10=1]

Solution:

According to the equation,

\; {\Ni}+2 {\Ag}^{+} → {\Ni}^{2+}+2 {\Ag}

\;∆ {\G}=-n {\FE}^{∘}

\;\;\text " where "\; \;\; ∆ {\G}= {\Gibb}^{'} \;\text " ' free energy "\;

\;∆ {\G}=-2 × 96500 × 1.05

\; {\N}=\;\text " No. of electrons gain or lost "\;=2

\;∆ {\G}=-202.650 {\kJ}

\; {\F}=\;\text " Faraday's constant "\;=96500

\; {\E}^{∘}=\;\text " Standard emf "\;=1.05 {\V}

The relation between Gibb's free energy and Equilibrium constant is given by equation

{{\E}^{∘}}_\cell =\;{0.0591}/{n} \log {\K}_c

\log {\K}_c\;=-\;{1.05 × 2}/{0.0591}=35.53

{\K}_c\;=3.39 × 10^{35}

OR
According to the equation,

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

{\E}°_{\;\text "cell "\;} \;= {\E}°_{\;\text "cathode "\;} - {\E}°_{\;\text "anode "\;}

{\E}°_{\;\text "cell "\;} =0-(-0.44) {\V}

{\E}°_{\;\text "cell "\;} =+0.44 {\V}

By applying Nernst Equation,

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

\; {\E}_{\;\text "cell "\;}=0.44 \;{-0.0591}/{2} \log \;{0.001}/{(0.01)^2}

\; {\E}_{\;\text "cell "\;}=0.44-\;{0.0591}/{2} \log 10

\; {\E}_{\;\text "cell "\;}=0.44-0.0295 × 1

\; {\E}_{\;\text "cell "\;}=+0.410 {\V}

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