03. ELECTROCHEMISTRY (HM)
Explanation:
कैथोड पर;
\(F{e^{2 + }} + 2{e^ - } \to Fe\) ; \(F{e^{3 + }} + 3{e^ - } \to Fe\)
\({({E_{Fe}})_1} = \frac{{{\rm{Atomic}}{\rm{.weight}}}}{2};\,\,{({E_{Fe}})_2} = \frac{{{\rm{Atomic}}{\rm{.weight}}}}{3}\)
मुक्त \(Fe\) के भार का अनुपात
\( = \frac{{{\rm{Atomic weight}}}}{3}:\,\frac{{{\rm{Atomic weight}}}}{2} = \,3\,\,:\,\,2\).