SELECT * FROM question_mgmt as q WHERE id=3237 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=2 AND subId=9 AND chapterId=49 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 11th chemistry Redox Reactions the mn3 ion is unstable in solution and undergoes

Question:

The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2, and H+ ion. Write a balanced ionic equation for the reaction.

Answer:

The given reaction can be represented as:

Mn3+(aq)  →  Mn2+(aq)  + MnO2(s)  +  H+(aq)

The oxidation half equation is:

M+3 n3+(aq)  →  M+4 nO2(s)

The oxidation number is balanced by adding one electron as:

Mn3+(aq)     →  M nO2(s) + e-

The charge is balanced by adding 4H+ ions as:

Mn3+(aq)     →  M nO2(s)  +  4H+(aq)   +    e-

The O atoms and H+ ions are balanced by adding 2H2O molecules as:

Mn3+(aq)   +  2H2O(l)   →    MnO2(s)  + 4H+(aq)   +    e..... (i)

The reduction half equation is:

Mn3+(aq)     →   Mn2+(aq)  

The oxidation number is balanced by adding one electron as:

Mn3+(aq)   +    e-    →   Mn2+(aq)   ....... (ii)

The balanced chemical equation can be obtained by adding equation (i) and (ii) as:

2Mn3+(aq)   +  2H2O(l)   →   MnO2(s)  + 2Mn2+(aq) + 4H+(aq)


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SELECT ex_no,question,question_no,id,chapter FROM question_mgmt as q WHERE courseId='2' AND subId='9' AND ex_no!=0 AND status=1 and id!=3237 ORDER BY last_viewed_on desc limit 0,10

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  • Answered by Ekta Mehta
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