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leonid [27]
3 years ago
6

Consider the half reactions below for a chemical reaction.

Chemistry
1 answer:
ladessa [460]3 years ago
7 0

Answer:

Option A:

Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)

Explanation:

The half reactions given are:

Zn(s) → Zn^(2+)(aq) + 2e^(-)

Cu^(2+) (aq) + 2e^(-) → Cu(s)

From the given half reactions, we can see that in the first one, Zn undergoes oxidation to produce Zn^(2+).

While in the second half reaction, Cu^(2+) is reduced to Cu.

Thus, for the overall reaction, we will add both half reactions to get;

Zn(s) + Cu^(2+) (aq) + 2e^(-) → Cu(s) + Zn^(2+)(aq) + 2e^(-)

2e^(-) will cancel out to give us;

Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)

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The standard reduction potentials for the Ag+|Ag(s) and Zn2+| Zn(s) half-cell reactions are +0.799 V and -0.762 V, respectively.
mihalych1998 [28]

<u>Answer:</u> The potential of the given cell is 1.551 V

<u>Explanation:</u>

The given chemical cell follows:

Zn(s)|Zn^{2+}(0.125M)||Ag^{+}(0.240M)|Ag(s)

<u>Oxidation half reaction:</u> Zn(s)\rightarrow Zn^{2+}(0.125M)+2e^-;E^o_{Zn^{2+}/Zn}=-0.762V

<u>Reduction half reaction:</u> Ag^{+}(0.240M)+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.799V       ( × 2)

<u>Net cell reaction:</u> Zn(s)+2Ag^{+}(0.240M)\rightarrow Zn^{2+}(0.125M)+2Ag(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

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

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

Putting values in above equation, we get:

E^o_{cell}=0.799-(-0.762)=1.561V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Zn^{2+}]}{[Ag^{+}]^2}

where,

E_{cell} = electrode potential of the cell = ? V

E^o_{cell} = standard electrode potential of the cell = +1.561 V

n = number of electrons exchanged = 2

[Zn^{2+}]=0.125M

[Ag^{+}]=0.240M

Putting values in above equation, we get:

E_{cell}=1.561-\frac{0.059}{2}\times \log(\frac{(0.125)}{(0.240)^2})

E_{cell}=1.551V

Hence, the potential of the given cell is 1.551 V

6 0
4 years ago
Please help!-- 20 pts!
ratelena [41]

Answer:

Both oil and gasoline molecules are nonpolar, while water is polar. Nonpolar solvents have a tendency to dissolve other nonpolar molecules.

Explanation:

Molecules may be categorized as "polar" or "nonpolar" according to <em>difference in the atom's electronegativity.</em>

<u>Water is polar</u> because it consists of two types of atoms that<em> do not cancel out each other.</em> It is made of two atoms of Hydrogen and only one atom of Oxygen. This makes the Oxygen<u> partially negative</u> and the Hydrogen <u>partially positive.</u> This allows them to readily bond with other polar molecules like sugar. However, it cannot mix freely with oil and gasoline because<em> both of these are nonpolar. </em>Nonpolar molecules do not have much difference when it comes to their atoms' electronegativity. <em>Therefore, they have the tendency to dissolve molecules which are nonpolar as well. </em>This explains why oil molecules can mix freely with gasoline.

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2 years ago
Is bubbles forming in water a chemical change
Stolb23 [73]
It is a physical change

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3 years ago
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I am not sure about the first question but the temperature has an important role in this situation because as the temp goes up particles moves at a faster speed and spread out every where.
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If you have 1kg of carbon and 1 kg of gold, they will weigh the same.
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This would be be true
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