Chromium is electroplated from an aqueous solution containing H2SO4 and H2CrO4. What current is required to deposit chromium at rate of 1.25g/min.
1.25g x 1mol/51.9961 = .0240mol
Now I just don't know how to figure out the number of electrons to do the rest of the steps
What is the oxidation state change in Cr starting in H2CrO4 to Cr as the element? That tells you how many electrons per Cr atom were required.
Is it 6 ?
If it is 6, then it should be
1.25g x 1mol/51.9961 = .0240mol
.0240mol x 6 mol e- =.144
.144mol e-/60sec x 96500 = 232
6 is correct.
232 WHAT? The number is ok. What unit?
Amperes
Thanks
The unit is A. I'm not sure what it means, but its just what my teacher taught us.
A represents amperes, which is the unit of electrical current. It is a measure of the rate at which electric charge flows through a circuit.
You're welcome! The current required to deposit chromium at a rate of 1.25g/min is 232 Amperes.
To calculate the current required to deposit chromium at a rate of 1.25g/min, you first need to determine the number of electrons needed to deposit 1 mole of chromium. The oxidation state change of chromium starting from H2CrO4 to Cr as the element is 6.
Now that you have determined the oxidation state change, you can proceed with the calculations. First, convert 1.25g of chromium to moles:
1.25g x (1 mol / 51.9961g) = 0.0240mol
Next, multiply the number of moles by the number of electrons per mole of chromium:
0.0240mol x 6 mol e- = 0.144 mol e-
Finally, convert from mole of electrons to seconds using Faraday's constant (96500 C/mol):
0.144mol e- x (1 C / 96500 mol e-) = 0.001494C
Therefore, the current required to deposit chromium at a rate of 1.25g/min is 0.001494C/sec, which can be expressed as 232 Amperes.