A current of 0.500 A flows through a cell containing Fe2+ for 10.0 minutes. Calc

Question

A current of 0.500 A flows through a cell containing Fe2+ for 10.0 minutes. Calculate the maximum moles of Fe that can be removed from solution? Assume constant current over time (Faraday constant = 9.649 x 104 C/mol).

A)

1.04 mmol

B)

51.8 mol

C)

3.11 mmol

D)

1.55 mmol

E)

25.9 mol

A current of 0.500 A flows through a cell containing Fe2+ for 10.0 minutes. Calculate the maximum moles of Fe that can be removed from solution? Assume constant current over time (Faraday constant = 9.649 x 104 C/mol).

A)

1.04 mmol

B)

51.8 mol

C)

3.11 mmol

D)

1.55 mmol

E)

25.9 mol

Explanation / Answer

Fe2+ + 2e- ----> Fe

We see from the reaction that to convert 1 mole of Fe2+ to Fe we need 2 moles of electrons

Given,

A current of 0.500 A flows through a cell containing Fe2+ for 10.0 minutes

Total charge passes = current x time (s)

=> Charge = 0.5 x 10 x 60 = 300 C

We know that,

Charge on 1 e- = 1.602 x 10^-19 C

=> Number of electrons = 300 / 1.602 x 10^-19 = 1.87 x 10^21

Moles of e- = Number of e- / Avagadro number

=> Moles of e- = 1.87 x 10^21 / 6.022 x 10^23 = 3.11 x 10^-3 moles

=> Moles of Fe that can be removed = 3.11 x 10^-3 / 2 = 1.55 x 10^-3 moles = 1.55 mmol

D) 1.55 mmol

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