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Consider the latimer diagram given below :

The \mathrm{E^0} in the given diagram is ______ V.

(Approximate the answer to 1 digit after decimal place e.g 6.7 or 0.7)

Option: 1

0.5


Option: 2

0.6


Option: 3

0.7


Option: 4

0.8


Answers (1)

best_answer

As we have learnt,

Gibbs energy of the reaction is given as 

\mathrm{\Delta_{r}G^0=-nFE^0_{cell}}

Now, using the \mathrm{ClO_3^- |ClO^-} couple, we can write the reaction as 

\mathrm{ClO _{3}^{-}+2 H _{2} O +4 e^{-} \rightarrow ClO ^{-}+4 OH ^{-}; \Delta G _{1}^{0}} \quad (1)

and, using the \mathrm{ClO^- |Cl^-} couple, we can write 

\mathrm{ClO^- + H_2O + 2e^- \longrightarrow Cl^- + 2OH^-; \Delta G^0_2} \quad (2)

Adding equations (1) and (2), we can write 

\mathrm{ClO_3^- + 3 H_2O + 6e^- \longrightarrow Cl^- + 6 OH^-; \Delta G_3^0}

Thus, the three Gibb's Energy change can be written as 

\mathrm{ \Delta G_3^0=\Delta G_1^0+\Delta G_2^0}

\mathrm{ -6FE_3^0=-4F(0.54)-2F(0.71)}

\mathrm{ E_3^0=\frac{4(0.54)+2(0.71)}{6}= 0.6 ~V}

Posted by

Ritika Kankaria

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