Equivilant Mass...?

Question

I did a redox experiment where I got H2 to be released from water by sending electricity into a metal in water. The metal lost .024 grams and 50 mL of H2(g) was collected.
I established the half reaction was 2e-+H2O->H2, and the problem said, (1 mole e- = 1 faraday)
The following problem says, "With knowledge of EM, the molar mass of your metal (MM) may be calculated using the equation MM = EM x n. Here, n is the charge of oxidaized metal cation, so MM simply represents a multiple of EM. For instac,e for EM=24, MM has possible values of 24 g mol^-1 (for n=+1), 48 gmol^-1 (for n=+2)...etc.

The problem is "The equivilant mass (EM) of a metal is the mass of metal oxidized by 1 f (faraday).In this experiment EM is calculated using the following equations
EM=metal lost(g)/f
This expression assumes that 1f is required to oxidize 1 EM of metal. Using your data, calculate the EM for your metal."

In order for that to happen, EM has to be above 1 but metal lost is .024, so how do I get EM>1?

Answer 1

Your half reaction: 2e-+H2O->H2, is not balanced.
It should be: 2H2O + 2e- ==> H2(g) + 2OH-
Also, please specify the temperature T and pressure P of H2(g) you collected. From the ideal gas formula, PV = nRT, you will be able to find n, the number of moles of H2. Since it takes 2 moles of electrons to be transferred to generate one mole of H2(g), you should have:
0.24/ (EM) = 2*n = 2PV/RT
Hence: (EM) = 0.24*RT/(2PV)
You do the math.