Question
The oxidation state of $I$ in ${H_4}IO_6^ - $ is
$4 + x - 12 = - \,1$; $x = - 1 + 8$$ = + \,7$.
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$(C= 3 \times 10^8 \,ms^{-1}$ and $N_A =6.02 \times mol^{-1}).$
$CH_3CH=CHCH_2CHBrCH_3$


$Cu ( s )+ Sn ^{2+}( aq ) \rightarrow Cu ^{2+}( aq )+ Sn ( s )$
$\left( E _{ Sn ^{2+} \mid Sn }^{0}=-0.16\, V , E _{ Cu ^{2+} \mid Cu }^{0}=0.34\, V \right.$ Take $F=96500\, C\, mol ^{-1}$ )