The equilibrium Cr_2O_7^{2-} + H_2O rightleft | vedclass

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(B) The given equilibrium is $Cr_2O_7^{2-} + H_2O \rightleftharpoons 2CrO_4^{2-} + 2H^{+}$.According to Le Chatelier's principle,if we add a base ($OH

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a basic medium

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(B) The given equilibrium is $Cr_2O_7^{2-} + H_2O \rightleftharpoons 2CrO_4^{2-} + 2H^{+}$.According to Le Chatelier's principle,if we add a base ($OH^-$ ions) to the system,the $OH^-$ ions will react with the $H^+$ ions produced in the reaction to form water $(H^+ + OH^- \rightarrow H_2O)$.This removal of $H^+$ ions from the product side decreases the concentration of $H^+$,which causes the equilibrium to shift to the right to produce more $H^+$ ions.Therefore,the equilibrium shifts to the right in a basic medium.

The equilibrium $Cr_2O_7^{2-} + H_2O \rightleftharpoons 2CrO_4^{2-} + 2H^{+}$ is shifted to the right in:

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