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OLEGan [10]
2 years ago
15

Whats the voltage of CuCl2 + Zn -> ZnCl2 + Cu

Chemistry
1 answer:
gtnhenbr [62]2 years ago
7 0

Answer:

Approximately 1.10\; {\rm V} under standard conditions.

Explanation:

Equation for the overall reaction:

{\rm CuCl_{2}}\, (aq) + {\rm Zn}\, (s) \to {\rm ZnCl_{2}} \, (aq) + {\rm Cu}\, (s).

Write down the ionic equation for this reaction:

\begin{aligned}& {\rm Cu^{2+}}\, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Zn}\, (s)\\ & \to {\rm Zn^{2+}} \, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Cu}\, (s)\end{aligned}.

The net ionic equation for this reaction would be:

{\rm Cu^{2+}}\, (aq) + {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + {\rm Cu}\, (s).

In this reaction:

  • Zinc loses electrons and was oxidized (at the anode): {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}.
  • Copper gains electrons and was reduced (at the cathode): {\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s).

Look up the standard potentials for each half-reaction on a table of standard reduction potentials.

Notice that {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} is oxidation and is likely not on the table of standard reduction potentials. However, the reverse reaction, {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s), is reduction and is likely on the table.

  • E(\text{anode}) = -0.7618\; {\rm V} for {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s), and
  • E(\text{cathode}) = 0.3419\; {\rm V} for {\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s).

The reduction potential of {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} would be -E(\text{anode}) = -(-0.7618\; {\rm V}) = 0.7618\; {\rm V}, the opposite of the reverse reaction {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s).

The standard potential of the overall reaction would be the sum of the standard potentials of the two half-reactions:

\begin{aligned} E^{\circ} &= E^{\circ}(\text{cathode}) + (-E^{\circ}(\text{anode})) \\ &= 0.3419 - (-0.7618\; {\rm V}) \\ &\approx 1.10\; {\rm V}\end{aligned}.

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3 years ago
A sodium hydroxide solution is made by mixing 8.70 g NaOH with 100 g of water. The resulting solution has a density of 1.087 g/m
Ahat [919]

Answer:

Mass fraction = 0.08004

Mole fraction = 0.0377

Explanation:

Given, Mass of NaOH = 8.70 g

Mass of solution = 8.70 + 100 g = 108.70 g

Mass\ fraction\ of\ NaOH=\frac {8.70}{108.70} = 0.08004

Molar mass of NaOH = 39.997 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{8.70\ g}{39.997\ g/mol}

Moles\ of\ NaOH= 0.2175\ mol

Given, Mass of water = 100 g

Molar mass of water = 18.0153 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{100\ g}{18.0153\ g/mol}

Moles\ of\ water= 5.5508\ mol

So, according to definition of mole fraction:

Mole\ fraction\ of\ NaOH=\frac {n_{NaOH}}{n_{NaOH}+n_{water}}

Mole\ fraction\ of\ NaOH=\frac {0.2175}{0.2175+5.5508}=0.0377

3 0
3 years ago
Having energy in the reactant lead to an exothermic reaction.<br>true<br>false​
Ostrovityanka [42]

Answer:

false

Explanation:

first of all;-energy lead to an indotermic reaction.

indotermic is a reaction that absorbs energy \

*it has positive enthalpy of reaction

*Heat content of product is greater than that of reactant

*Heat is added to reactant side

example;- CO^2+2H^2+891kj --------- CH4 +2O2

6 0
3 years ago
The combustion of 0.25 mol of an unknown organic compound results in the release of 320 kJ of energy. Which of the compounds in
vichka [17]

Answer: C. ethanol

The enthalpy of combustion is the amount of heat produced when one mole of ethanol undergoes complete combustion at 25 ° C and 1 atmosphere pressure, yielding products also at 25 ° C and 1 atm.

<u>The enthalpy of combustion of the unknown compound is</u>

ΔH = - 320 kJ / 0.25 mol = - 1280 kJ / mol

<u>To choose a probable compound according to this combustion enthalpy, we must evaluate the deviation in relation to the values reported in the literature for the three probable compounds</u> (methane, ethylene and ethanol). The deviation (e%) will be calculated according to the following equation,

e% =  ( | ΔHx - ΔH | / ΔHx ) x 100%

where ΔHx is the enthalpy of combustion of the probable compound.

The following table shows the combustion enthalpies of the probable compounds and their deviation in relation to the enthalpy of ΔH = - 1280 kJ / mol

Compound Enthalpy of combustion (kJ/mol)   Deviation

Methane                        - 890.7                                 43.8%

Ehylene                         -1411.2                                   9.3%

Ethanol                        -1368.6                                    6.5%

According to the previous table, we can say that the most probable compound is ethanol, since it has the smallest deviation in relation to the experimental enthalpy value of combustion.

5 0
3 years ago
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