cr2+ is reduced to cr(s) at an electrode surface. If a current of 50 amperes is

Question

cr2+ is reduced to cr(s) at an electrode surface. If a current of 50 amperes is maintained for 11.0 hours, what mass of chromium is deposited on the electrode? Assume 100% current efficiency. (F= 96500 C/mol e^-) cr2+ is reduced to cr(s) at an electrode surface. If a current of 50 amperes is maintained for 11.0 hours, what mass of chromium is deposited on the electrode? Assume 100% current efficiency. (F= 96500 C/mol e^-) cr2+ is reduced to cr(s) at an electrode surface. If a current of 50 amperes is maintained for 11.0 hours, what mass of chromium is deposited on the electrode? Assume 100% current efficiency. (F= 96500 C/mol e^-)

Explanation / Answer

Solution :-

Given data

Time = (11.0 hours * 60 min / 1 hr )*(60 sec / 1 min ) =39600 sec

Current = 50 A

Mass of Chromium deposited= ?

We are given with the current and time

Therefore using the current and time lets first calculate the charge in coloumb (C)

Charge = current * time

Lets put the values in the formula

Charge (C) = 50 A * 39600 sec

                     = 1980000 C

Now using the charge lets calculate the moles of electrons

1 mole e- = 96500 C

? mol e- = 1980000 C

1980000 C * 1 mol e- / 96500 C = 20.52 mol e-

Now using the moles of electrons lets calculate the moles of the Cr deposited.

Cr^2+ + 2e-   -------- > Cr(s)

1 mol Cr = 2 mol e-

20.52 mol e- * 1 mol Cr / 2 mol e- = 10.26 mol Cr

Now lets convert moles of Cr to its mass

Mass = moles * molar mass

Mass of Cr= 10.26 mol * 51.996 g per mol

                   = 533 g Cr

Therefore mass of the Cr deposited on the electrode = 533 g

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