What is the value of n in the following equation :
`Cr(OH)_(4)^(-)+OH^(-) rarr CrO_(4)^(2-)+H_(2)O+n e`?

What is the value of n in the following half equation, Cr(OH)_(4)^(-) + OH^(-) rarrCrO_(4)^(2-)+H_(2)O+"ne"^(-)

In the following reaction: Cr(OH)_(3)+OH^(-)+IO_(3)^(-) toCrO_(4)^(2-)+H_(2)O+I^(-)

The values of coefficients to balance the following reaction are Cr(OH)_(3)+ClO^(-)+OH^(-) rarr CrO_(4)^(2-)+Cl^(-)+H_(2)O

The value of n in the following half equation is MnO4^(-) + 2H_2O + ne rarr MnO_2 + 4O'H

Complete the following equation : (i) Cr2O_(7)^(2-)+2OH^(-)rarr (ii) MnO_(4)^(-)+4H^(+)+3e^(-)rarr

How many moles of O_(2) will be liberated by one mole of CrO_(5) is the following reaction: CrO_(5) + H_(2) SO_(4) rarr Cr_(2) (SO_(4))_(5) + H_(2)O + O_(2)

How many moles of O_(2) will be liberated by one mole of CrO_(5) is the following reaction: CrO_(5) + H_(2) SO_(4) rarr Cr_(2) (SO_(4))_(5) + H_(2)O + O_(2)