Math  /  Algebra

QuestionWhen the following skeletal equation is balanced under basic conditions, what are the coefficients of the species shown? \square N2H4+\mathrm{N}_{2} \mathrm{H}_{4}+ \square SnO22\mathrm{SnO}_{2}{ }^{2-} \qquad , \square NH3+\mathrm{NH}_{3}+ \square SnO32\mathrm{SnO}_{3}{ }^{2-}
Water appears in the balanced equation as a \square (reactant, product, neither) with a coefficient of \square . (Enter 0 for neither.)
Which species is the oxidizing agent? \square

Studdy Solution

STEP 1

What is this asking? We need to balance a chemical equation in a basic solution, figure out the coefficients for each chemical, determine if water is a reactant, product, or neither, and identify the oxidizing agent. Watch out! Balancing in basic solutions adds an extra layer of complexity compared to acidic solutions.
Don't forget those hydroxide ions!

STEP 2

1. Assign oxidation numbers
2. Write half-reactions
3. Balance half-reactions
4. Combine half-reactions
5. Add hydroxide ions

STEP 3

Let's **assign oxidation numbers** to each element in the equation to see what's changing.
In N2H4\mathrm{N}_{2} \mathrm{H}_{4}, nitrogen has an oxidation number of 2-2 and hydrogen has +1+1.
In SnO22\mathrm{SnO}_{2}{ }^{2-}, tin has +2+2 and oxygen has 2-2.
In NH3\mathrm{NH}_{3}, nitrogen has 3-3 and hydrogen has +1+1.
Lastly, in SnO32\mathrm{SnO}_{3}{ }^{2-}, tin has +4+4 and oxygen has 2-2.

STEP 4

Now, let's **write the half-reactions**.
Nitrogen goes from 2-2 in N2H4\mathrm{N}_{2} \mathrm{H}_{4} to 3-3 in NH3\mathrm{NH}_{3}, gaining one electron per nitrogen atom, or two electrons per N2H4\mathrm{N}_{2} \mathrm{H}_{4} molecule: N2H42NH3+2e \mathrm{N}_{2} \mathrm{H}_{4} \rightarrow 2\mathrm{NH}_{3} + 2e^{-} Tin goes from +2+2 in SnO22\mathrm{SnO}_{2}{ }^{2-} to +4+4 in SnO32\mathrm{SnO}_{3}{ }^{2-}, losing two electrons: SnO22SnO32+2e \mathrm{SnO}_{2}{ }^{2-} \rightarrow \mathrm{SnO}_{3}{ }^{2-} + 2e^{-}

STEP 5

Next, we **balance the half-reactions**.
The nitrogen half-reaction needs four hydrogen ions on the right side: N2H42NH3+2e+2H+ \mathrm{N}_{2} \mathrm{H}_{4} \rightarrow 2\mathrm{NH}_{3} + 2e^{-} + 2\mathrm{H}^{+} The tin half-reaction needs one oxygen atom on the left.
We add water to balance oxygen and hydrogen ions to balance hydrogen: SnO22+H2OSnO32+2e+2H+ \mathrm{SnO}_{2}{ }^{2-} + \mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{SnO}_{3}{ }^{2-} + 2e^{-} + 2\mathrm{H}^{+}

STEP 6

Since both half-reactions involve two electrons, we can **combine them** directly: N2H4+SnO22+H2O2NH3+SnO32+2H++2e+2e+2H+ \mathrm{N}_{2} \mathrm{H}_{4} + \mathrm{SnO}_{2}{ }^{2-} + \mathrm{H}_{2}\mathrm{O} \rightarrow 2\mathrm{NH}_{3} + \mathrm{SnO}_{3}{ }^{2-} + 2\mathrm{H}^{+} + 2e^{-} + 2e^{-} + 2\mathrm{H}^{+} Notice that the electrons and hydrogen ions add to zero on both sides, simplifying to: N2H4+SnO22+H2O2NH3+SnO32 \mathrm{N}_{2} \mathrm{H}_{4} + \mathrm{SnO}_{2}{ }^{2-} + \mathrm{H}_{2}\mathrm{O} \rightarrow 2\mathrm{NH}_{3} + \mathrm{SnO}_{3}{ }^{2-}

STEP 7

Since this is a basic solution, we don't want hydrogen ions.
We **add hydroxide ions** to both sides to neutralize them: N2H4+SnO22+H2O2NH3+SnO32 \mathrm{N}_{2} \mathrm{H}_{4} + \mathrm{SnO}_{2}{ }^{2-} + \mathrm{H}_{2}\mathrm{O} \rightarrow 2\mathrm{NH}_{3} + \mathrm{SnO}_{3}{ }^{2-} This reaction is already balanced for a basic solution.

STEP 8

The balanced equation is N2H4+SnO22+H2O2NH3+SnO32\mathrm{N}_{2} \mathrm{H}_{4} + \mathrm{SnO}_{2}{ }^{2-} + \mathrm{H}_{2}\mathrm{O} \rightarrow 2\mathrm{NH}_{3} + \mathrm{SnO}_{3}{ }^{2-}.
The coefficients are **1, 1, 1, 2, and 1**.
Water is a **reactant** with a coefficient of **1**.
The oxidizing agent is SnO22\mathrm{SnO}_{2}{ }^{2-} because tin gains electrons (it oxidizes something else).

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