1 $0 \mathrm{~L}$ each of a buffer containing $1 \mathrm{~mole} \mathrm{NH}_{3}$ and $1 \mathrm{~mol}$ of…

1 $0 \mathrm{~L}$ each of a buffer containing $1 \mathrm{~mole} \mathrm{NH}_{3}$ and $1 \mathrm{~mol}$ of $\mathrm{NH}_{4}^{+}$ were placed in the cathodic and anodic half-cells and $965 \mathrm{C}$ of electricity was passed. If anodic and cathodic half cells reactions involve oxidation and reduction of water only as
$2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 4 \mathrm{H}^{+}+\mathrm{O}_{2}-4 \mathrm{e}^{-}$
$2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-}$
Then $\mathrm{pH}$ of
  1. cathodic solution will increase
  2. anodic solution will decrease
  3. both the solutions will remain practically constant
  4. both the solutions will increase

Solution

Due to buffer action the $\mathrm{pH}$ will remain practically constant. ,

Asked in: JEE-TOPICTESTS-CHEMISTRY

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