During the electrolysis of concentrated $H_2SO_4$,perdisulphuric acid $(H_2S_2O_8)$ and $O_2$ are formed at the anode in equimolar amounts. The moles of $H_2$ that will form simultaneously at the other electrode will be (Given: $2H_2SO_4 \rightarrow H_2S_2O_8 + 2H^+ + 2e^-$)
- Athrice of $O_2$
- Btwice of $O_2$
- Cequal to $O_2$
- Dhalf of $O_2$
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$A$ negative $e.m.f.$ for a cell represents:
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View SolutionThe molar conductance of $NaCl$,$HCl$,and $CH_{3}COONa$ at infinite dilution are $126.45$,$426.16$,and $91.0 \ S \ cm^{2} \ mol^{-1}$ respectively. The molar conductance of $CH_{3}COOH$ at infinite dilution is. Choose the right option for your answer. (In $S \ cm^{2} \ mol^{-1}$)
At $300 \ K$,the $E_{cell}^{\circ}$ of $A_{(s)} + B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + B_{(s)}$ is $1.0 \ V$. If $\Delta_r S^{\circ}$ of this reaction is $100 \ J \ K^{-1} \ mol^{-1}$,what is $\Delta_r H^{\circ}$ (in $kJ \ mol^{-1}$) of this reaction? $(F = 96500 \ C \ mol^{-1})$
Match 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 $pH = 10$ | $(I)$ $0.76 \ V$ |
| $(B)$ $Cu^{2+}|Cu$ | $(II)$ $0.059$ |
| $(C)$ $Zn|Zn^{2+}$ | $(III)$ $-0.591 \ V$ |
| $(D)$ $\frac{2.303RT}{F}$ | $(IV)$ $0.337 \ V$ |
| $(V)$ $-0.76 \ V$ |
$(a)$ $A-III, B-I, C-II, D-V$
$(b)$ $A-II, B-V, C-I, D-IV$
$(c)$ $A-III, B-IV, C-I, D-II$
$(d)$ $A-V, B-I, C-IV, D-II$
Given that $E^o_{Fe^{2+}/Fe} = -0.44 \ V$,$E^o_{Cu^{2+}/Cu} = 0.34 \ V$,and $E^o_{Ag^+/Ag} = 0.80 \ V$. Which of the following statements is correct?
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