(a) ${{Cr}_{2}O}_{7}^{2 -} + 14H^{+} + 6I^{-} \longrightarrow 2{Cr}^{3 +} + 7H_{2}O + 3I_{2}$ ${Cr}_{2}O_{7}^{2 -}$is reduced to${Cr}^{3 +}.$ Thus, final state of Cr is +3. Hence, (a)
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