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Question

The standard electrode potential $$(E^{\ominus})$$ for $$OCl^-|Cl^-$$ and $$Cl^-|\dfrac{1}{2}Cl_2$$ respectively are $$0.94$$ V and $$-1.36$$ V. The $$E^{\ominus}$$ value of $$OCl^-|\dfrac{1}{2}Cl_2$$ will be:


A
0.42 V
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B
2.20 V
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C
0.52 V
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D
1.04 V
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Solution

The correct option is C $$0.52$$ V
$$ E^{\circ} $$ for $$ O\bar{Cl}/\bar{Cl}\Rightarrow 0.94\,v $$
$$ E^{\circ} $$ for $$ \bar{Cl}/1/2C_{12}\Rightarrow -1.36\,v $$
$$ O\bar{Cl}\rightarrow \bar{Cl}---(1) \rightarrow \Delta C1^{0}, n = 3 $$
$$ \bar{Cl}\rightarrow \dfrac{1}{2}Cl_{2}---(2)\rightarrow \Delta C2^{0} $$  $$ n = 1 $$
$$ (1)+(2) $$
$$ O\bar{Cl}\rightarrow 1/2Cl_{2}--(3)\rightarrow \Delta C^{0} $$  $$ n = 1 $$
So
$$ \Delta C^{0} = \Delta C_{1}^{0}+\Delta C_{2}^{0} $$
$$ -nFE^{0} = -nFE_{1}^{0}-nFE_{2}^{0} $$ 
$$ -1\times F\times E^{0} = -3\times F\times 0.94-1.36\times 1\times F$$  
So $$[E^{0} = 0.52\,v] $$

1199760_1037255_ans_bd9642871f7644759a30a800ba7799fe.jpg

Chemistry

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