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nataly862011 [7]
3 years ago
6

The galvanic cell illustrated above generates a potential of +1.10V. For the construction of a second galvanic cell (not shown),

only one modification was made: the Cu electrode has double the mass of the Cu electrode in the first cell. Which of the following correctly compares the initial E° for the second cell to that of first cell at 298K, and why?
Chemistry
1 answer:
xxTIMURxx [149]3 years ago
5 0

Answer:

The initial  E°  for the second cell is the same as for the first cell.

Explanation:

Because the overall chemical reaction that occurs in the cell does not change.

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P = 11.133 atm (purple)

T = -236.733 °C(yellow)

n = 0.174 mol(red)

<h3>Further explanation  </h3>

Some of the laws regarding gas, can apply to ideal gas (volume expansion does not occur when the gas is heated),:  

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\large {\boxed {\bold {PV = nRT}}}

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V = volume, liter  

n = number of moles  

R = gas constant = 0.08206 L.atm / mol K  

T = temperature, Kelvin  

To choose the formula used, we refer to the data provided

Because the data provided are temperature, pressure, volume and moles, than we use the formula PV = nRT

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T= 10 +273.15 = 373.15 K

V=5.5 L

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\tt P=\dfrac{nRT}{V}\\\\P=\dfrac{2\times 0.08205\times 373.15}{5.5}\\\\P=11.133~atm

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V=8.3 L

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T = 12 + 273.15 = 285.15 K

V=3.4 L

P=1.2 atm

\tt n=\dfrac{PV}{RT}\\\\n=\dfrac{1.2\times 3.4}{0.08205\times 285.15}\\\\n=0.174~mol

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