Polymers — Chemistry STD 12 Science — Question

$0.48 + \log \left\{ { - \frac{{d[A]}}{{dt}}} \right\} = \log \left\{ { + \frac{{d[B]}}{{dt}}} \right\} + 0.7$

Then, $x : y$ is

Above reaction is called

${B^{3 + }}_{(aq.)} + 3{e^ - } \to B(s)$  ; $E = 0.6\ Volt$

Using above information find the ion which will be deposited first at cathode if solution containing $A_{(aq.)}^{2 + }$ & $B_{(aq.)}^{ 3+ }$ is electrolysed

${[Fe\,{(CN)_6}]^{4 - }}\, \to \,{[Fe{(CN)_6}]^{3 - }}\, + \,{e^ - }\,;\,$ ${E^o}\, = \, - \,0.35\,V$

$\,F{e^{2 + }}\, \to \,F{e^{3 + }}\, + \,{e^ - }\,;$                                 ${E^o}\, = \, - \,0.77\,V$