This is for my chemistry test. I have posted a sample from my Chemistry teacher.
ID: 929400 • Letter: T
Question
This is for my chemistry test. I have posted a sample from my Chemistry teacher. Can someone tell me what formula "he is using". Please print your answer SO THAT I MAY COPY AND PASTE IT. Do not utilize different notations. He gives partial credit on his test but only if I am using the same notation as him. Thus expalin what the Q means and the n and the F. Also do the computation so as to acquire the same solutions within the picture. Also, label your notation to the side.
10 pts 6. 6 pts a) How many moles of metal Mo is plated out of a solution containing MoO4 if a constant current of 550 milliamps is used in the electrolysis and the process takes 12.00-min??+-+Mw04" Mo + 4H2 2- . 376 C b) 4 pts If the anode compartment of the above cell contains a Pb(s) electrode which is oxidized to PbSO4 (s), under the above conditions, how many moles of PbSO4 (s) are formed? 2t-Explanation / Answer
In the above question the formula used is:
Q = nFN(moles)
F is faraday constant and is denoted by F and has a constant value which is 96488 C mol-1
n is number of moles electrons transferred.
Q is charge on n moles of electrons and is given by Q = nF
Q is quantity of charge flowing through the electrolyte can be calculated from current strength and time for which current is passed.
therefore , Q = I * t
Q = quantity of charge
I = Current in ampheres
T = time in seconds.
A) In the above questions we have to find the moles of metal Mo
we can use the formula: Q = nFN(mol)
N (mol) = Q / nF
6e-+ 8H+ + MoO42- ---> Mo + 4H2O ( n = 6 because 6 moles of electrons are used in equation)
F(faraday constant) = 96488 C mol-1 (constant value )
now we can find the Q by using formula Q = I*T
I = 550 miliamps(given in question)
we have to change miliamps into amphere:
1 miliamps = 0.001 amphere
550 miliamps = (0.001amphere) * (550 miliamps) / (1 miliamps) = 0.55 amphere
T = 12 min, we have to change it into seconds
1 min = 60 seconds
12 minutes = 12 * 60 seconds = 720 seconds
I = 0.55 amphere T = 720 seconds
Q = I * T
Q = 0.55 amphere * 720 seconds
Q = 396 C
Now we have the values of Q , n and F now substitude these values in equation : Q = nFN(mol)
396 C = 6 (96488) N (mol)
N (mol) = 396C / 6( 96488)
N(mol) = 0.000684 = 6.84 * 10-4 mol
b) Same method we will use in this question as we used in previous part.
By using formula : Q = nFN(mol)
Q is same which is 396C
F(faraday constant) = 96488 C mol-1 (constant value )
Reaction : SO42- + Pb ---> PbSO42- + 2e- ( n= 2)
substitude the values of Q , n and F in equation
Q = nFN(mol)
396C = 2(96488) N(mol)
N(mol) = 396 / 2(96488)
N(mol) = 0.00205 = 2.05 * 10-3 mol
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