CBSE Class 12 Chemistry 2017 Outside Delhi Set 1 Solved Paper

Question : 12 of 26

Marks: +1, -0

(a) The cell in which the following reaction occurs:

2 {\Fe}^{3+}( {\aq})+2 {\I}^{-}( {\aq}) → 2 {\Fe}^{2+}( {\aq})

+ {\I}_2( {\ s})

has

{\E}°_{\;\text "cell "\;} =0.236 {\V}

298 {\K}

. Calculate the standard Gibb's energy of the cell reaction.
(Given :

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

)
(b) How many electrons flow through a metallic wire if a current of

0.5 {\A}

is passed for 2 hours ?
(Given :

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

)

Solution:

\;\text " (a) "\;

\;2 {\Fe}^{3+}+2 {\e}^{-} → 2 {\Fe}^{2+}

2 {\I}^{-} → {\I}_2+2 {\e}^{-}

For the given cell reaction,

n=2

∆ {\G}^{∘}=-n {\FE}°_{\;\text "cell "\;}

\;=-2 × 96500 × 0.236

\;=-45548 {\J} {\mol}^{-1}

\;=-45.55 {\kJ} {\mol}^{-1}

(b)

{\I}=0.5^{ {\A} t=2 \;\text " hours "\;\;=2 × 60 × 60 s=7200 {\ s}}

{\Q}\;= {\I} t

\;=0.5 × 7200

\;=3600 \;\text " coulombs "\;

A flow of

96500 {\c}

is equal to flow of 1 mole of electrons is

6.023 × 10^{23}

electrons.

∴ 3600 {\c}

is equivalent to of electrons

\;=\;{6.023 × 10^{23}}/{96500} × 3600

\;=2.246 × 10^{22} \;\text " electrons "\;

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