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The number of electrons required to reduce chromium completely in $\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}$ to $\mathrm{Cr}^{3+}$ in acidic medium, is
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6
$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}$
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