NTA Abhyas 2020
- Exam
- NTA Abhyas
- Year
- 2020
- Subject
- Chemistry
- Chapter
- Redox Reactions
NTA Abhyas 2020 · Chemistry · Redox Reactions
An equal volume of a reducing agent is titrated separately with 1 M...
An equal volume of a reducing agent is titrated separately with 1 M KMnO4 in acid, neutral and alkaline medium. The volumes of KMnO4 required are 20 mL in acid, 33.3 mL in neutral and 100 mL in alkaline media. Find out the volume (mL) of 1M K2Cr2O7 consumed, if the same volume of the reducing agent is titrated in acid medium.
Report your answer by rounding it up to nearest whole number.
Report your answer by rounding it up to nearest whole number.
Correct answer
17
Step-by-step solution
Let equal volumes of R.A in each case be X Meqs
\begin{cases}{\text{MnO}_{4}^{-} + 8 {\text{H}}^{+} + 5 {\text{e}}^{-} & \to & {\text{Mn}}^{2 +} + 4 {\text{H}_{2} \text{O} & \left(\text{Acid}\right) & \left(\text{Z = 5}\right) \\ {\text{MnO}_{4}^{-} + 2 {\text{H}_{2} \text{O} + 3 {\text{e}}^{-} & \to & {\text{MnO}_{2} + 4 {\text{OH}}^{-} & \left(\text{Neutral}\right) & \left(\text{Z = 3}\right) \\ {\text{MnO}_{4}^{-} + {\text{e}}^{-} & \to & {\text{MnO}_{4}^{2 -} & \text{Alkaline} & \left(\text{Z = 1}\right)\end{cases}
\left(\right)\text{=}\left(\right)\text{=}\left(\right)\text{=}\left(\right)\text{=100 milli}\text{eqs}{\text{Cr}_{2} {\text{O}_{7}^{2 -} + 1 4 {\text{H}}^{+} + 6 {\text{e}}^{-} \to 2 {\text{Cr}}^{3 +} + 7 {\text{H}_{2} \text{O}{\text{Cr}_{2} {\text{O}_{7}^{2 -} = 6 \times 1 \times \text{Volume mL,}$
Vol = 16.7 mL
\begin{cases}{\text{MnO}_{4}^{-} + 8 {\text{H}}^{+} + 5 {\text{e}}^{-} & \to & {\text{Mn}}^{2 +} + 4 {\text{H}_{2} \text{O} & \left(\text{Acid}\right) & \left(\text{Z = 5}\right) \\ {\text{MnO}_{4}^{-} + 2 {\text{H}_{2} \text{O} + 3 {\text{e}}^{-} & \to & {\text{MnO}_{2} + 4 {\text{OH}}^{-} & \left(\text{Neutral}\right) & \left(\text{Z = 3}\right) \\ {\text{MnO}_{4}^{-} + {\text{e}}^{-} & \to & {\text{MnO}_{4}^{2 -} & \text{Alkaline} & \left(\text{Z = 1}\right)\end{cases}
\left(\right)\text{=}\left(\right)\text{=}\left(\right)\text{=}\left(\right)\text{=100 milli}\text{eqs}{\text{Cr}_{2} {\text{O}_{7}^{2 -} + 1 4 {\text{H}}^{+} + 6 {\text{e}}^{-} \to 2 {\text{Cr}}^{3 +} + 7 {\text{H}_{2} \text{O}{\text{Cr}_{2} {\text{O}_{7}^{2 -} = 6 \times 1 \times \text{Volume mL,}$
Vol = 16.7 mL