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4 years ago

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Oxygen, nitrogen, and fluorine bond with hydrogen to form molecules. These molecules are attracted to each other by

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andre [41]4 years ago

4 0

Answer:

The oxygen, nitrogen, and fluorine bonds with hydrogen to produce molecules, these molecules are fascinated towards each other by hydrogen bonds.

A hydrogen bond refers to a partially electrostatic attraction amid a hydrogen atom that is combined with a more electronegative group or atoms like oxygen, nitrogen, or fluorine. In this bonding, one is known as the hydrogen bond donor, and another atom adjacent to it exhibiting a lone pair of electrons is known as the hydrogen bond

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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.

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<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

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