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
B
2.20 V
C
0.52 V
D
1.04 V

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 $$[E^{0} = 0.52\,v]$$Chemistry

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