Question

==>What is disproportionation reaction?? Give example
==>calculate the pH of 0.05M of Ba(OH)2 solution...
Pls ans fast.....
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Answer 1

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➤ A disproportionation reaction is when a multiatomic species whose pertinent element has a specific oxidation state gets oxidized and reduced in two separate half-reactions, yielding two other products containing the same pertinent element. A convenient example is Mn2O3 becoming Mn2+ and MnO2 .

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➤ pH value of the resulting solution is 12.

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Answer 2

Explanation:

==>What is disproportionation reaction?? Give example

==>calculate the pH of 0.05M of Ba(OH)2 2 Al0(s) + 3 Cl20(g) --> 2 Al+3Cl3-1(s)

Al is oxidized and Cl in Cl2 is reduced. There are 6 electrons involved.

(b) 8 H+(aq) + MnO4-(aq) + 5 Fe2+(aq) --> 5 Fe3+(aq) + Mn2+ (aq) + 4 H2O(l)

8 H+(aq) + [Mn+7O4-2 ]-(aq) + 5 Fe2+(aq) --> 5 Fe3+(aq) + Mn2+ (aq) + 4 H2+O-2(l)

Fe2+ is oxidized, Mn in MnO4- is reduced. There are 5 electrons involved.

(c) FeS(s) + 3 NO3-(aq) + 4 H+(aq) --> 3 NO(g) + SO42-(aq) + Fe3+(aq) + 2 H2O(l)

Fe+2S-2(s) + 3 [N+5O3-2]-(aq) + 4 H+(aq) --> 3 N+2O-2(g) + [S+6O4-2]2-(aq) + Fe3+(aq) + 2 H2+O-2(l)

Fe in FeS is oxidized, S in FeS is oxidized.

N in NO3- is reduced. There are 9 electrons involved.

(d) 10 H+(aq) + 8 I-(aq) + SO42-(aq) --> H2S(aq) + 4 I2(s) + 4 H2O(l)

10 H+(aq) + 8 I-(aq) + [S+6O4-2]2-(aq) --> H2+S-2(aq) + 4 I20(s) + 4 H2+O-2(l)

I- is oxidized, S in SO42- is reduced. There are 8 electrons involved.

(e) H2O2(aq) + 2 Fe2+(aq) + 2 H+(aq) --> 2 Fe3+(aq) + 2 H2O(l)

H2+O2-(aq) + 2 Fe2+(aq) + 2 H+(aq) --> 2 Fe3+(aq) + 2 H2+O-2(l)

Fe2+ is oxidized. O in H2O2 is reduced. There are 2 electrons involved.

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