Answer : The cell potential for this reaction is 0.50 V
Explanation :
The given cell reactions is:
The half-cell reactions are:
Oxidation half reaction (anode): 
Reduction half reaction (cathode): 
First we have to calculate the cell potential for this reaction.
Using Nernest equation :
![E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B2.303RT%7D%7BnF%7D%5Clog%20%5Cfrac%7B%5BZn%5E%7B2%2B%7D%5D%7D%7B%5BPb%5E%7B2%2B%7D%5D%7D)
where,
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature = 
n = number of electrons in oxidation-reduction reaction = 2
= standard electrode potential of the cell = +0.63 V
= cell potential for the reaction = ?
![[Zn^{2+}]](https://tex.z-dn.net/?f=%5BZn%5E%7B2%2B%7D%5D)
= 3.5 M
![[Pb^{2+}]](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D)
= 
Now put all the given values in the above equation, we get:
Therefore, the cell potential for this reaction is 0.50 V
Add 7 water atom to the right hand side to adjust the quantity of oxygen. Increase Cr(+3) by two to adjust the quantity of Cr. Duplicate Cl-by two to adjust the quantity of chlorine molecules.
Cr2O7[2-](aq) +2 Cl[-](aq) < - >2 Cr[3+] (aq) + Cl2(g)+7H2O
Presently adjust that charges.
you have - 4 charges on the left hand side, while +18 charges on the right hand side, there for include 14H+ the left hand side to adjust the charges
Cr2O7[2-](aq) +2 Cl[-](aq)+14H+ < - >2 Cr[3+] (aq) + Cl2(g)+7H2O
take note of that the oxidation number of hydrogen in water is +1
Answer:
C) solid, liquid, gas
Explanation:
Which grouping shows a decrease in Intermolecular
Forces of Attraction?
A) gas, liquid, solid B) liquid, solid, gas
C) solid, liquid, gas D) solid, gas, liquid
the further the particles are from each other, the less the intermolecular attraction they are farthest in a gas, then a liquid, and closest in a solid
Oxygen,hydrogen, and carbon
