The standard electrode potentials $E^o_{(I_2/I^{-})}$,$E^o_{(Br^{-}/Br_2)}$,and $E^o_{(Fe/Fe^{2+})}$ are respectively $+0.54 \ V$,$-1.09 \ V$,and $0.44 \ V$. On the basis of the above data,which of the following processes is non-spontaneous?
- A$Br_2 + 2I^{-} \to 2Br^{-} + I_2$
- B$Fe + Br_2 \to Fe^{2+} + 2Br^{-}$
- C$Fe + I_2 \to Fe^{2+} + 2I^{-}$
- D$I_2 + 2Br^{-} \to 2I^{-} + Br_2$
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Similar Questions
An aqueous solution containing $6.5 \ g$ of $NaCl$ of $90 \%$ purity was subjected to electrolysis. After the complete electrolysis,the solution was evaporated to get solid $NaOH$. The volume of $1 \ M$ acetic acid required to neutralize $NaOH$ obtained above is (in $cm^{3}$)
MediumKCET 2009
View SolutionMatch the following:
List-$I$ List-$II$ $(A)$ Potential of hydrogen electrode at $pH = 10$ $(I)$ $0.76 \ V$ $(B)$ $Cu^{2+}|Cu$ $(II)$ $0.059$ $(C)$ $Zn|Zn^{2+}$ $(III)$ $-0.591 \ V$ $(D)$ $\frac{2.303RT}{F}$ $(IV)$ $0.337 \ V$ $(V)$ $-0.76 \ V$
$(a)$ $A-III, B-I, C-II, D-V$
$(b)$ $A-II, B-V, C-I, D-IV$
$(c)$ $A-III, B-IV, C-I, D-II$
$(d)$ $A-V, B-I, C-IV, D-II$
| List-$I$ | List-$II$ |
|---|---|
| $(A)$ Potential of hydrogen electrode at $pH = 10$ | $(I)$ $0.76 \ V$ |
| $(B)$ $Cu^{2+}|Cu$ | $(II)$ $0.059$ |
| $(C)$ $Zn|Zn^{2+}$ | $(III)$ $-0.591 \ V$ |
| $(D)$ $\frac{2.303RT}{F}$ | $(IV)$ $0.337 \ V$ |
| $(V)$ $-0.76 \ V$ |
$(a)$ $A-III, B-I, C-II, D-V$
$(b)$ $A-II, B-V, C-I, D-IV$
$(c)$ $A-III, B-IV, C-I, D-II$
$(d)$ $A-V, B-I, C-IV, D-II$
Choose the correct representation of conductometric titration of benzoic acid vs sodium hydroxide.
DifficultJEE MAIN 2023
View SolutionMolar conductances of $BaCl_2$,$H_2SO_4$,and $HCl$ at infinite dilutions are $x_1, x_2$,and $x_3$,respectively. Equivalent conductance of $BaSO_4$ at infinite dilution will be
Difficult
View SolutionConsider the change in oxidation state of Bromine corresponding to different $emf$ values as shown in the diagram below:
$BrO_4^{-}$ $\xrightarrow{1.82 \ V} BrO_3^{-}$ $\xrightarrow{1.5 \ V} HBrO$ $\xrightarrow{1.0652 \ V} Br_2$ $\xrightarrow{1.595 \ V} Br^{-}$
Then the species undergoing disproportionation is:
$BrO_4^{-}$ $\xrightarrow{1.82 \ V} BrO_3^{-}$ $\xrightarrow{1.5 \ V} HBrO$ $\xrightarrow{1.0652 \ V} Br_2$ $\xrightarrow{1.595 \ V} Br^{-}$
Then the species undergoing disproportionation is:
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