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Nitella [24]
3 years ago
13

In the presence of a dielectric, the capacitance of a electric field inside the plates now is: a) Less b) More c) Same as the el

ectric field in absence of the dielectric d) Zero
Physics
1 answer:
Nesterboy [21]3 years ago
6 0

Answer:

Explanation:

As the dielectric is inserted between the plates of a capacitor, the capacitance becomes K times and the electric field between the plates becomes 1 / K times the original value. Where, K be the dielectric constant.

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Missing data in the text of the exercise: The molar concentration of Zinc is 10 times the molar concentration of copper.

Solution:

1) First of all, let's calculate the standard electrode potential difference at standard temperature. This is given by:
E^0=E_{cat}^0-E_{an}^0
where E_{cat}^0 is the standard potential at the cathode, while E_{an}^0 is the standard potential at the anode. For a Daniel Cell, at the cathode we have copper: E_{Cu}^0=+0.34 V, while at the anode we have zinc: E_{Zn}^0=-0.76 V. Therefore, at standard temperature the electrode potential difference of the Daniel Cell is
E^0=+0.34 V-(-0.76 V)=+1.1 V

2) To calculate E^0 at any temperature T, we should use Nerst equation:
E^0(T)=E^0- \frac{R T}{z F} \ln  \frac{[Zn]}{[Cu]}
where 
R=8.31 J/(K mol)
T=473.15 K is the temperature in our problem
z=2 is the number of electrons transferred in the cell's reaction
F=9.65\cdot 10^4 C/mol is the Faraday's constant
[Zn] and [Cu] are the molar concentrations of zinc and in copper, and in our problem we have [Zn]=10[Cu].
Using all these data inside the equation, and using E^0=+1.1 V, in the end we find:
E^0(T)=E^0- \frac{R T}{z F} \ln \frac{[Zn]}{[Cu]}=+1.053 V
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