Answer: 0.18 V
Explanation:-
Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.
=-0.40V[/tex]
=-0.24V[/tex]
Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.
Where both 
are standard reduction potentials.
![E^0=E^0_{[Ni^{2+}/Ni]}- E^0_{[Cd^{2+}/Cd]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BNi%5E%7B2%2B%7D%2FNi%5D%7D-%20E%5E0_%7B%5BCd%5E%7B2%2B%7D%2FCd%5D%7D)
Using Nernst equation :
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cd^{2+}]}{[Ni^{2+]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BCd%5E%7B2%2B%7D%5D%7D%7B%5BNi%5E%7B2%2B%5D%7D)
where,
n = number of electrons in oxidation-reduction reaction = 2
= standard electrode potential = 0.16 V
![E_{cell}=0.16-\frac{0.0592}{2}\log \frac{[0.10]}{[0.5]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D0.16-%5Cfrac%7B0.0592%7D%7B2%7D%5Clog%20%5Cfrac%7B%5B0.10%5D%7D%7B%5B0.5%5D%7D)
Thus the potential of the following electrochemical cell is 0.18 V.
Answer:
The answer to your question is V2 = 5.09 l
Explanation:
Data
Volume 1 = V1 = 5.0 L
Temperature 1 = T1 = 5°C
Volume 2 = V2 = ?
Temperature 2 = T2 = 10°C
Formula (Charles law)
V1/T1 = V2/T2
-Solve for V2
V2 = V1T2 / T1
-Convert temperature to °K
T1 = 5 + 273 = 278°K
T2 = 10 + 273 = 283°K
-Substitution
V2 = (5)(283) / (278)
-Simplification
V2 = 1415 / 278
-Result
V2 = 5.09 l
From high temperature to low temperature.
Answer:
a. 1 mole of acid is equal to one equivalent.
b. 1.00 moles of HCl are found.
c. 1L of 2.00M NaOH is needed to reach the equivalence point
Explanation:
HCl reacts with NaOH as follows:
HCl + NaOH → NaCl + H2O
<em>Where 1 mole of HCl reacts with 1 mole of NaOH. The reaction is 1:1</em>
a. As the reaction is 1:1, 1 mole of acid is equal to one equivalent
b. The initial moles of HCl are:
1.00L * (2.00moles HCl / 1L) = 2.00 moles of HCl
At the halfway point, the moles of HCl are the half, that is:
1.00 moles of HCl are found
c. At equivalence point, we need to add the moles of NaOH needed for a complete reaction of the moles of HCl. As the moles of HCl are 2.00 and the reaction is 1:1, we need to add 2.00 moles of NaOH, that is:
2.00moles NaOH * (1L / 2.00mol) =
1L of 2.00M NaOH is needed to reach the equivalence point
