A quantity of electrical charge that brings about the deposition of 4.5 g Al from Al3+ at a cathode will also produce what volume (STP) of H2(g) from H+ at a cathode?

I know there are 3 e- difference b/w the two reactions. I took 4.5 g/3 and get 1.5 g H2. Then (1.5 g H2)(1 mol H2/2 g H2)(22.4 L/mol H2)=16.8 L. Answer in the book says 5.6 L ? What am I doing wrong? Thanks.

96,485 coulombs (1 Faraday) will deposit 1 equivalent weight of Al (27/3 = 9 grams).

Your problem has 4.5 g; therefore, there must have been 1/2 x 96,485 coulombs (1/2 F). That same quantity of electricity will release 1 equivalent weight of hydrogen or H. That is 1/2 of a mole (1/2 of H2) and a mole occupies 22.4L @ STP; therefore, 1/2 mole will occupy 1/2 x 22.4 = ?
Atoms and ions react as moles and not grams.

To find the volume of H2 gas produced from H+ at the cathode, you need to use the stoichiometry of the reaction and the ideal gas law.

First, let's determine the number of moles of Al deposited. We are given that the deposition of 4.5 g of Al occurs via the reduction of Al3+ ions. The molar mass of Al is approximately 26.98 g/mol, so the number of moles of Al deposited is:

moles of Al = mass of Al / molar mass of Al
= 4.5 g / 26.98 g/mol
≈ 0.1669 mol

Since there is a 3:2 ratio between the number of moles of electrons transferred and the number of moles of H2 produced, we can calculate the number of moles of H2 gas formed using this ratio:

moles of H2 = (3/2) * moles of Al
≈ (3/2) * 0.1669 mol
≈ 0.2503 mol

Now, we can use the ideal gas law to calculate the volume of H2 at STP (Standard Temperature and Pressure). The ideal gas law is expressed as:

PV = nRT

Where:
P = pressure, which is 1 atm at STP
V = volume, which is what we are trying to find
n = number of moles of gas
R = ideal gas constant, which is 0.0821 L·atm/(mol·K)
T = temperature, which is 273 K at STP

Rearranging the equation and solving for V:

V = (n * R * T) / P
= (0.2503 mol * 0.0821 L·atm/(mol·K) * 273 K) / 1 atm
≈ 5.6942 L

Rounding to the appropriate number of significant figures, the volume of H2 gas produced at STP is approximately 5.6 L.

Based on your calculations, it seems like you made a mistake when converting the mass of H2 from grams to moles. The molar mass of H2 is approximately 2 g/mol, not 1 g/mol. So, the correct conversion factor should be (1 mol H2 / 2 g H2). Using the correct conversion, you should get the same answer of approximately 5.6 L.