- AForm sulphur halides
- BAre oxidised
- ✓Are reduced
- DNone of these
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Precipitate $X$ is
$(A)$ $Fe _4\left[ Fe ( CN )_6\right]_3$
$(B)$ $Fe \left[ Fe ( CN )_6\right]$
$(C)$ $K _2 Fe \left[ Fe ( CN )_6\right]$
$(D)$ $KFe \left[ Fe ( CN )_6\right]$
Among the following, the brown ring is due to the formation of
$(A)$ $\left[ Fe ( NO )_2\left( SO _4\right)_2\right]^{2-}$
$(B)$ $\left[ Fe ( NO )_2\left( H _2 O \right)_4\right]^{3+}$
$(C)$ $\left[ Fe ( NO )_4\left( SO _4\right)_2\right]$
$(D)$ $\left[ Fe ( NO )\left( H _2 O \right)_5\right]^{2+}$

$(i)$ ${C_4}{H_5}N{H_2}$ $(ii)$ $C{H_3}N{H_2}$
$(iii)$ ${(C{H_3})_2}NH$ $ (iv)$ ${(C{H_3})_3}N$

$Pt ( s )\left| H _2( s )( latm )\right| H ^{+}\left( aq ,\left[ H ^{+}\right]=1\right)|| Fe ^{3+}( aq ), Fe ^{2+}( aq ) \mid \operatorname{Pt}( s )$
Given : $E _{ Fe ^{3+} / e ^{2 *}}^0=0.771\,V$ and $E _{ H ^{+}+\frac{1}{2} H _2}^0=0 V , T =298\,K$
If the potential of the cell is $0.712\,V$ the ratio of concentration of $Fe ^{2+}$ to $Fe ^{2+}$ is $........$.(Nearest integer)