(c) CrO 4 2 ? ( aq ) + S 2 O 3 2 ? ( aq ) ? Cr 3+ ( aq ) + SO 4 2 ? ( aq ) (d) A
ID: 818727 • Letter: #
Question
(c) CrO42?(aq) + S2O32?(aq) ? Cr3+(aq) + SO42?(aq)
(d) Al(s) + MnO4?(aq) ? MnO2(s) + Al(OH)4?(aq)
(e) NO2?(aq) + Al(s) ? NH3(g) + AlO2?(aq)
Balance the following oxidation-reduction reactions, which occur in basic solution, using the half-reaction method. (Use the lowest possible coefficients. Include states-of-matter under the given conditions in your answer.)(a) MnO4?(aq) + S2?(aq) ? MnS(s) + S(s) Cl2(g) ? Cl ?(aq) + ClO?(aq) CrO42?(aq) + S2O32?(aq) ? Cr3+(aq) + SO42?(aq) Al(s) + MnO4?(aq) ? MnO2(s) + Al(OH)4?(aq) NO2?(aq) + Al(s) ? NH3(g) + AlO2?(aq)Explanation / Answer
(B) Half reactions means separate oxidation and reduction reactions
Cl2(g) + 2OH-(aq) ---> ClO-(aq) + Cl-(aq) + H2O(l)
In this Cl2 is undergoing self oxidation and reduction
Cl2(g)-----> ClO-(aq)(oxidation)
Cl2(g) ----> Cl-(aq) (reduction)
(C)Let's assume the reaction occurs in a base solution.
If so, we first balance O by adding OH-
and next balance H by adding H2O.
Separate the reaction into 2 half-reactions.
CrO42- -----> Cr3+
SO32- -----> SO42-
Balance the main atoms first (Cr and S).
They are already balanced in this reaction.
Balance the oxygen by adding OH-.
CrO42- -----> Cr3+ + 4OH-
SO32- + OH- -----> SO42-
Now balance the H by adding H2O and adjusting OH-.
CrO42- + 4H2O -----> Cr3+ + 8OH-
SO32- + 2OH- -----> SO42- + H2O
Next, add e- to the more positive side of
each half-reaction to conserve charge.
CrO42- + 4H2O + 3e- -----> Cr3+ + 8OH-
SO32- + 2OH- -----> SO42- + H2O + 2e-
To make the number of e- gained and lost equal,
multiply the first equation by 2 and the second by 3.
2CrO42- + 8H2O + 6e- -----> 2Cr3+ + 16OH-
3SO32- + 6OH- -----> 3SO42- + 3H2O + 6e-
Finally, add the two half reactions together,
simplifying the number of H2O and OH-.
2CrO42- + 3SO32- + 5H2O -----> 2Cr3+ + 3SO42- + 10OH-
(D) Al(s) + MnO4-(aq) --> MnO2(s) + Al(OH)4-(aq)
Al + 4OH- --> Al(OH)4^- + 3e- ......................oxidation
2H2O + MnO4- + 3e- --> MnO2 + 4OH- ........reduction
----------------------------- -----------------------------
Al + MnO4- + 2H2O --> Al(OH)4- + MnO2
Cl2 --> Cl-(aq) + ClO-(aq) ..... disproportionation reaction
Cl2 + 2e- --> 2Cl-
Cl2 + 2H2O --> 2OCl- + 4H+ + 2e-
------------------ -----------------------
2Cl2 + 2H2O --> 2Cl- + 2OCl- + 4H+
simplify
Cl2 + H2O --> Cl- + OCl- + 2H+
(E) NO2 + 7 H+ + 7e- = NH3 + 2 H2O) x 4
Al + 2 H2O = AlO2 + 4 H+ + 4e- ) x 7
4 NO2 + 7 Al + 6 H2O = 4 NH3 + 7 AlO2
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