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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Chemistry
Chapter
Redox Reactions

NTA Abhyas 2020 · Chemistry · Redox Reactions

What is the value of n in the following half equation, Cr(OH)_ 4 ^...

What is the value of n in the following half equation,

{\text{Cr(OH)}_{4}^{−} \text{+}{\text{OH}^{−} \to {\text{CrO}_{4}^{\text{2} -} \text{+}{H}_{2} \text{O}\text{+}{\text{ne}^{−} \text{?}

Options

  1. A3
  2. B6
  3. C5
  4. D2

Correct answer

A. 3

Step-by-step solution

Cr(OH)4−+OH−→CrO42−+H2OCr ( OH{)}_{4}^{-} + OH^{-} \to Cr{O}_{4}^{2 -} + {H}_{2}O

To balance H and O atoms, add 3H2O3{H}_{2}O to R.H.S. and 30H⊖30H^{⊖} to the L.H.S. we have,

Cr(OH)4−+4OH−→CrO42−+4H2OCr ( OH{)}_{4}^{-} + 4OH^{-} \to Cr{O}_{4}^{2 -} + 4{H}_{2}O

To balance charge, add 3e−3e^{-} to the R.H.S. we have

Cr(OH)4−+4OH−→CrO42−+4H2O+3e−Cr ( OH{)}_{4}^{-} + 4OH^{-} \to Cr{O}_{4}^{2 -} + 4{H}_{2}O + 3e^{-}
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