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4 years ago

8

Can someone tell me the answer ?

1 answer:

ludmilkaskok [199]4 years ago

7 0

The Aufbau principle states that, hypothetically, electrons orbiting one or more atoms fill the lowest available energy levels before filling higher levels (e.g., 1s before 2s). In this way, the electrons of an atom, molecule, or ion harmonize into the most stable electron configuration possible.

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3 years ago

A 75 gram bar of gold is cut into three equal pieces. How does the density of each piece compare to the density of the original

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4 years ago

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alexira [117]

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4 years ago

A wind with a direction designated as 90° is blowing from the ____

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3 years ago

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