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Electrochemical Cells, Galvanic Cells

The chemical ...

Question

The chemical reaction given below:
$C l_{2} (g) + S O_{2} (g) + 2 H_{2} O (l) \to 2 C l^{-} (a q .) + 3 H^{+} (a q .) + H S O_{4}^{-} (a q .)$
proceeds readily in aqueous acid solution.
(i) Give the half-cell reactions.
(ii) If a fully charged cell initially held $1$ mole of $C l_{2}$ , for how many days could it sustain a current of $0.05$ ampere, assuming the cell becomes non-operative when $90$ % of initial $C l_{2}$ has been used up?

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Solution

$C l_{2} + 2 H_{2} O + S O_{2} \to 2 C l^{-} + 3 H^{+} + H S O_{4}^{-}$
$C l_{2} r e d u c e d i n t o 2 C l^{-}$
$S O_{2} o x i d i s e d i n t o H S O_{4}^{-}$
$H a l f c e l l r e a c t i o n$
$(a) S O_{2} + 2 H_{2} O \to 3 H^{+} + H S O_{4}^{-} + 2 e^{-}$
$(b) C l_{2} + 2 e^{-} \to 2 C l^{-}$
$g i v e n : i = 0.05 A$
$n_{f} = 2$
$m o l . w t . o f C l_{2} = 70.9$
$n = 1 m o l$
$m = Z i t$
$m o l e \times m o l . w t = Z i t = \frac{m o l . w t \times i t}{n_{f} \times F}$
$0.90 \times 1 \times 70.9 = \frac{70.9 \times 0.05 \times t}{2 \times 96500}$
$t = 3474000 s e c$
$t = 40.2083 d a y s$
$K C l \to C l_{2} + K O H$
$m = \frac{M o l . W t .}{n F} \times Q$
$m = \frac{70.9}{2 \times 96500} \times 10000$
$m a s s o f C l_{2} = 3.67 g$
$d e n s i t y o f C l_{2} A t S T P = 2.994 \frac{g r a m}{L}$
$V o l u m e O f C l_{2} = \frac{3.67}{2.994}$
$= 1.22578 L$
$f o r H_{2}$
$m = \frac{2}{2 \times 96500} \times 10000$
$m a s s o f H_{2} = 0.10362 g$
$d e n s i t y o f H_{2} = 0.083 \frac{g r a m}{L}$
$V o l u m e O f H_{2} = \frac{0.103}{0.083}$
$= 1.23 L$

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