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Vlada [557]
3 years ago
13

In a particular redox reaction, no2– is oxidized to no3– and cu2 is reduced to cu . Complete and balance the equation for this r

eaction in acidic solution. Phases are optional.
Chemistry
2 answers:
lianna [129]3 years ago
8 0

Reduction half reaction: Cu²⁺(aq) + 2e⁻ → Cu⁰(s).


Oxidation half reaction: NO₂⁻(aq) + H₂O(l) → NO₃⁻(aq) + 2H⁺(aq) + 2e⁻.  

Balanced chemical reaction:

Cu²⁺(aq) + NO₂⁻(aq) + H₂O(l) → Cu(s) + NO₃⁻(aq) + 2H⁺(aq).

Copper is reduced from oxidation number +2 (Cu²⁺) to oxidation number 0 (Cu) and nitrogen is oxidized from oxidation number +3 (in NO₂⁻) to oxidation number +5 (in NO₃⁻).


crimeas [40]3 years ago
8 0

Answer:

Cu²⁺(aq) + 3NO₂⁻(aq) ⇄ Cu⁰(aq) + 2NO₃⁻(aq)

Explanation:

A redox reaction is a reaction that occurs with oxidation and a reduction. The compound or the element that oxides are called reducted agent because it promotes the reduction of the other one. The compound or the element that reduces is called the oxidation agent, for the same reason.

In the oxidation, the substance increases its oxidation number: it loses electrons. The opposite occurs in reduction: there is a decrease in the oxidation number and the substance gain electrons.

So, the oxidation reaction is:

NO₂⁻(aq) ⇄ NO₃⁻(aq) + 2e⁻

And the reduction:

Cu²⁺(aq) + 2e⁻ ⇄ Cu⁰(aq)

Note that in the first one, the oxidation number of O is always -2, and the oxidation number of N was from +3 to +5.

The balanced reaction will be:

Cu²⁺(aq) + 3NO₂⁻(aq) ⇄ Cu⁰(aq) + 2NO₃⁻(aq)

The number of atoms of an element must be the same on each side of the equation.

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How much 6.0 m hno3 is needed to neutralize 39ml of 2 m koh
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Answer:

13mL

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

HNO3 + KOH —> KNO3 + H2O

From the balanced equation above, we obtained the following data:

Mole ratio of the acid (nA) = 1

Mole ratio of the base (nB) = 1

Step 2:

Data obtained from the question.

This includes the following:

Molarity of the acid (Ma) = 6M

Volume of the acid (Va) =?

Volume of the base (Vb) = 39mL

Molarity of the base (Mb) = 2M

Step 3:

Determination of the volume of the acid.

Using the equation:

MaVa/MbVb = nA/nB, the volume of the acid can be obtained as follow:

MaVa/MbVb = nA/nB

6 x Va / 2 x 39 = 1/1

Cross multiply to express in linear form

6 x Va = 2 x 39

Divide both side by 6

Va = (2 x 39)/6

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