1. Tardigrade
  2. Question
  3. Chemistry
  4. What is the value of n in the following half equation on balancing ? Cr(OH)-4 + OH- arrow CrO2-4 + H2O + ne-

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. What is the value of n in the following half equation on balancing ? $Cr(OH)^-_4 + OH^- \rightarrow CrO^{2-}_4 + H_2O + ne^-$

Redox Reactions

Solution:

$Cr ( OH )_{4}^{-}+ OH ^{(-)} \longrightarrow CrO _{4}^{2-}+ H _{2} O + ne ^{(-)}$
Reduction $\rightarrow 3 H _{2}^{+1} O _{2}^{-1}+6 e ^{(-)}+6 H _{2} O \longrightarrow 6^{+1} H _{2}^{-2} O +6 OH ^{(-)}$
So, Value of $n=6$