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Cr⁢2O⁢2−7+14⁢H⁡++6e⁢−->2______ + 7H2O - Chemistry (Theory)

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Question

\[\ce{Cr2O^{2-}_7 + 14H+ + 6e- -> 2}\]______ + 7H2O

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Solution

\[\ce{Cr2O^{2-}_7 + 14H+ + 6e- -> 2\mathbf{\underline{Cr^{3+}}}}\] + 7H2O

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