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Zolol [24]
3 years ago
14

In the galvanic cell Al(s) ǀ Al3+(aq, 1 M) ǀǀ Cu2+(aq, 1 M) ǀ Cu(s) which of the following changes will increase the cell potent

ial? I. Dilution of the Al3+ solution to 0.001 M II. Dilution of the Cu2+ solution to 0.001 M III. Increasing the surface area of the Al(s) electrode (A) I only (B) II only (C) III only (D) I and III only
Chemistry
2 answers:
cestrela7 [59]3 years ago
4 0

Answer:

(B) II only

Explanation:

dangina [55]3 years ago
3 0
The cathode electrode is copper. This means that the concentration of copper must be increased to increase the cell potential. This can be done by diluting the Al3+ solution. Increasing the surface area will not change the current but will increase the voltage. The answer is (D) I and III only.
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The amount of W(OH)2 needed would be 448.126 g

<h3>Stoichiometric calculation</h3>

From the equation of the reaction:

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The mole ratio of W(OH)2 to HCl is 1:2

Mole of 150g HCl = 150/36.461

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Equivalent mole of W(OH)2 = 4.11/2

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Mass of 2.06 moles W(OH)2 = 2.06 x 217.855

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More on stoichiometric calculations can be found here: brainly.com/question/8062886

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