MCQ
Iodoform is formed on warming ${I_2}$ and $NaOH$ with
- ✓${C_2}{H_5}OH$
- B$C{H_3}OH$
- C$HCOOH$
- D${C_6}{H_6}$
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$Zn^{2+} + 2e^-$ $\longrightarrow$ $Zn (s) ; E^o = -0.76\,V$
$Ca^{2+} + 2e^-$ $\longrightarrow$ $Ca (s) ; E^o = -2.87\,V$
$Mg^{2+} + 2e^-$ $\longrightarrow$ $Mg (s) ; E^o = -2.36\,V$
$Ni^{2+} + 2e^-$ $\longrightarrow$ $Ni (s) ; E^o = -0.25\,V$
The reducing power of the metals increases in the order
$(a)$ $NCl_3$ is formed when $Cl_2$ is present in excess.
$(b)$ $N_2$ is formed when $Cl_2$ is present in excess.
$(c)$ $N_2$ is formed when $NH_3$ is present in excess.
$(d)$ $NCl_3$ is formed when $NH_3$ is present in excess.
Correct option are
Statement $I$ : The higher oxidation states are more stable down the group among transition elements unlike p-block elements.
Statement $II$ : Copper can not liberate hydrogen from weak acids.
In the light of the above statements, choose the correct answer from the options given below :
$ {I_2} + 2{e^ - } \to \,2{I^ - }\,;\,\,{E^o}\, = \,\,0.54\,\,V $
$ MnO_4^ - \, + \,8{H^ + }\, + \,5{e^ - } \to \,M{n^{2 + }}\, + \,4{H_2}O\,;\,{E^{o\,}} = 1.52\,\,V $
$ F{e^{3 + }} + {e^ - } \to \,\,F{e^{2 + }}\,;\,\,{E^o}\, = \,\,0.77\,\,V $
$ S{n^{4 + }} + 2{e^ - } \to \,\,S{n^{2 + }}\,;\,\,{E^o}\, = \,\,0.1\,\,V $
The strongest reducant and oxidant respectively are