Solution:
At the point when Cr2O72–is treated with an antacid:
\[\left( orange \right)\text{ }Cr2O72+\text{ }OH-\to \text{ }2CrO42-\left( yellow \right)\]
At the point when the yellow arrangement is treated with a corrosive, we get back the orange arrangement:
\[\left( yellow \right)2CrO42-+2H+\to Cr2O72\left( orange \right)+\text{ }H2O~\]