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Annette [7]
3 years ago
6

Which of the following is true regarding a voltaic (or galvanic) cell? a) It can only be used with hydrogen. b) It produces elec

trical current spontaneously. c) It consumes electrical current to drive a nonspontaneous chemical reaction.
Chemistry
1 answer:
N76 [4]3 years ago
3 0

Answer:

b) It produces electrical current spontaneously.

Explanation:

Cells capable of converting chemical energy to electrical energy and vice versa are termed Electrochemical cells. There are two types of electrochemical cells viz: Galvanic or Voltaic cells and Electrolytic cells. Voltaic cell is an elctrochemical cell capable of generating electrical energy from the chemical reaction occuring in it.

The voltaic cell uses spontaneous reduction-oxidation (redox) reactions to generate ions in a half cell that causes electric currents to flow. An half cell is a part of the galvanic cell where either oxidation or reduction reaction is taking place. Hence, the spontaneous production of electric currents is true about Voltaic/Galvanic cells.

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One mole of an ideal gas is contained in a cylinder with a movable piston. The temperature is constant at 77°C. Weights are remo
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Answer:

The total work is 4957.45J

Explanation:

For an ideal gas, at constant temperature the definition of work (W) is

W = - P.dV = - n.R.T \int\limits^i_f {\frac{dV}{V} \\W = - n.R.T. Ln (\frac{V_{f}}{V_{I}})\\W = - n.R.T. Ln (\frac{P_{i}}{P_{f}})

where P is the pressure, V the volume, n the moles number, T the temperature and R the gas constant.

To solve the problem is necessary to replace the two steps in the equation

Stape 1: n  = 1 mol, R = 0.082atm.L/K.mol, T = 77ºC = 350K, Pi = 5.50atm and Pf = 2.43atm.

W_{1} = - 1molx0.082\frac{atm.L}{K.mol}x350KxLn (\frac{5.50atm}{2.43atm}) = 23.44atm.Lx101.325\frac{J}{atm.L} =2375.44J

Stape 2: n  = 1 mol, R = 0.082atm.L/K.mol, T = 77ºC = 350K, Pi = 2.43atm and Pf = 1.00atm.

W_{2} = - 1molx0.082\frac{atm.L}{K.mol}x350KxLn (\frac{2.43atm}{1.00atm}) = 25.48atm.Lx101.325\frac{J}{atm.L} =2582.01J

The total work is the sum of the two steps

W = W_{1} + W_{2} = 2375.44J + 2582.01J = 4957.45J

4 0
3 years ago
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