The following reaction occurs in acidic medium: $KMnO_4 + 8 H^{+} + 5 e^{-} \longrightarrow K^{+} + Mn^{2+} + 4 H_2O$. What is the equivalent weight of $KMnO_4$? (Molecular weight of $KMnO_4 = 158$)
- A$79$
- B$31.6$
- C$158$
- D$39.5$
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Permanganate $(VII)$ ion,$MnO_4^-$,in basic solution oxidises iodide ion,$I^-$,to produce molecular iodine $(I_2)$ and manganese $(IV)$ oxide $(MnO_2)$. Write a balanced ionic equation to represent this redox reaction.
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View SolutionDiscuss the types of redox reactions in detail.
$H_2O_2$ reduces $MnO_4^-$ ion to:
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$5CO(g) + I_2O_5(s) \longrightarrow 5CO_2(g) + I_2(s)$
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