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Question

Calculate the number of millimoles and milliequivalents of $${Cr_{2}O_{7}}^{2-}$$ ion in acid medium when $$100\ mL$$ of $$0.01\ M\ {Cr_{2}O_{7}}^{2-}$$ is reduced to $$Cr^{3+}$$ by $$Fe^{2+}$$.


Solution

$$\underset {1\ mol}{Cr_{2}O_{7}^{2-}} + 14H^{+} + \underset {6\ mol}{6e^{-}} \rightarrow 2Cr^{3+} + 7H_{2}O$$
$$\therefore 0.01\ M\ Cr_{2}O_{7}^{2-} \equiv 0.06\ N\ Cr_{2}O_{7}^{2-}$$
Number of millimoles $$= M\times V = 0.01\times 100 = 1$$
Number of milliequivalents $$= N\times V = 0.06\times 100 = 6$$.

Chemistry

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