1. Tardigrade
  2. Question
  3. Chemistry
  4. Cr2O2-7+X xrightarrow H+ Cr3++H2O + oxidised product of X . X in the above reaction cannot be

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. $Cr_{2}O^{2-}_{7}+X \xrightarrow{ H^{+} }Cr^{3+}+H_{2}O$ + oxidised product of $X$ . $X$ in the above reaction cannot be

Redox Reactions

Solution:

$Cr_{2}O^{2-}_{7}+ X \xrightarrow{ H^{+} }Cr^{3+}+H_{2}O$ + oxidised product of $X$ In the given reaction $X$ is oxidised and hence acting as reducing agent. Out of given choices $SO^{2-}_{4}$ cannot act as reducing agent, because sulphur is present in its highest oxidation state, $+6$