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3 years ago

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g You observed the formation of several precipitates in the Reactions in Solution lab exercise. Identify the precipitate in each

of the following reactions: a. The yellow precipitate formed in the reaction between KI and Pb(NO3)2 is . b. The white precipitate formed in the reaction between BaCl2 and H2SO4 is . c. The brown precipitate formed in the reaction between NaOH and FeCl3 is . d. The blue precipitate formed in the reaction between CuSO4 and NaOH is .

1 answer:

RUDIKE [14]3 years ago

6 0

<u>Answer:</u>

<u>For a:</u> Lead iodide is a yellow precipitate.

<u>For b:</u> Barium sulfate is a white precipitate.

<u>For c:</u> Ferric hydroxide is a brown precipitate.

<u>For d:</u> Copper (II) hydroxide is a blue precipitate.

<u>Explanation:</u>

Precipitation reaction is defined as the reaction where a solid precipitate (solid substance) is formed at the end of the reaction. It is insoluble in water.

For the given options:

<u>For (a):</u>

The chemical reaction between KI and lead (II) nitrate follows:

$2KI(aq)+Pb(NO_3)_2(aq)\rightarrow PbI_2(s)+2KNO_3(aq)$

The iodide of lead is generally insoluble in water. Thus, lead iodide is a yellow precipitate.

<u>For b:</u>

The chemical reaction between barium chloride and sulfuric acid follows:

$BaCl_2(aq)+H_2SO_4(aq)\rightarrow BaSO_4(s)+2HCl(aq)$

The sulfate of barium is insoluble in water. Thus, barium sulfate is a white precipitate.

<u>For c:</u>

The chemical reaction between NaOH and ferric chloride follows:

$3NaOH(aq)+FeCl_3(aq)\rightarrow Fe(OH)_3(s)+3NaCl(aq)$

The hydroxide of iron is insoluble in water. Thus, ferric hydroxide is a brown precipitate.

<u>For d:</u>

The chemical reaction between NaOH and copper sulfate follows:

$CuSO_4+2NaOH\rightarrow Cu(OH)_2+Na_2SO_4$

The hydroxide of copper is insoluble in water. Thus, copper (II) hydroxide is a blue precipitate.

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Consider the reaction of metallic copper with iron(!!) to give copper(ll) and ironin 0.77V Fe* (aq) + e-Fe (aq) Cup (aq) + 2e --

frosja888 [35]

Answer :

(a) The anode and cathode will be $E^o_{(Cu^{2+}/Cu)}$

and $E^o_{(Fe^{3+}/Fe^{2+})}$ respectively.

(b) The emf of cell potential is 1.022 V

Explanation :

(a) The standard reduction potentials for iron and copper are:

$E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

From the standard reduction potentials we conclude that, the substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

So, iron will undergo reduction reaction will get reduced. Copper will undergo oxidation reaction and will get oxidized.

The given cell reactions are:

Oxidation half reaction (anode): $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction (cathode): $Fe^{3+}+1e^-\rightarrow Fe^{2+}$

Thus, the anode and cathode will be $E^o_{(Cu^{2+}/Cu)}$

and $E^o_{(Fe^{3+}/Fe^{2+})}$ respectively.

(b) Now we have to calculate the potential of a cell.

Oxidation half reaction (anode): $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction (cathode): $Fe^{3+}+1e^-\rightarrow Fe^{2+}$

In order to balance that electrons, we will multiple the reduction reaction by 2, we get:

Oxidation half reaction (anode): $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction (cathode): $2Fe^{3+}+2e^-\rightarrow 2Fe^{2+}$

The overall cell reaction will be,

$2Fe^{3+}+Cu\rightarrow Cu^{2+}+2Fe^{2+}$

$E^o_{[Fe^{3+}/Fe^{2+}]}=2\times 0.77V=1.54V$

$E^o_{[Cu^{2+}/Cu]}=0.34V$

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o=E^o_{[Fe^{3+}/Fe^{2+}]}-E^o_{[Cu^{2+}/Cu]}$

$E^o=1.54V-(0.34V)=1.20V$

Now we have to calculate the cell potential.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}]^2[Cu^{2+}]}{[Fe^{3+}]^2}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = emf of the cell = ?

Now put all the given values in the above equation, we get:

$E_{cell}=1.20-\frac{0.0592}{2}\log \frac{(0.20)^2(0.25)}{(0.0001)^2}$

$E_{cell}=1.022V$

Therefore, the emf of cell potential is 1.022 V

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