MCQ
Prussian blue is formed when
- AFerrous sulphate reacts with $FeC{l_3}$
- ✓Ferric sulphate reacts with ${K_4}\left[ {Fe{{(CN)}_6}} \right]$
- CFerrous ammonium sulphate reacts with $FeC{l_3}$
- DAmmonium sulphate reacts with $FeC{l_3}$
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$F{e^{3 + }}(aq) + {e^ - } \to F{e^{ - 1}}(aq);\,{E^o} = + 0.77\,V$
$A{l^{3 + }}(aq) + 3{e^ - } \to Al(s);\,{E^o} = - 1.66\,V$
$B{r_2}(aq) + 2{e^ - } \to 2B{r^ - }(aq);\,{E^o} = + 1.08\,V$
Based on the data given above, reducing power of $F{e^{2 + }},\,Al$ and $B{r^ - }$ will increase in the order
Reason : The colour arises due to charge transfer.
