Question

Suppose 550.mmol of electrons must be transported from one side of an electrochemical cell to another in 49.0 minutes. Calculate the size of electric current that must flow.Be sure your answer has the correct unit symbol and round your answer to 3 significant digits.

Answer 1

Answer:

18.0 Ampere is the size of electric current that must flow.

Explanation:

Moles of electron , n = 550 mmol = 0.550 mol

1 mmol = 0.001 mol

Number of electrons = N

$N=N_A\times n$

Charge on N electrons : Q

$Q = N\times 1.602\times 10^{-19} C$

Duration of time charge allowed to pass = T = 49.0 min = 49.0 × 60 seconds

1 min = 60 seconds

Size of current : I

$I=\frac{Q}{T}=\frac{N\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}$

$=\frac{n\times N_A\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}$

$I=\frac{0.550 mol\times 6.022\times 10^{23} mol^{-1}\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}=18.047 A\approx 18.0 A$

18.0 Ampere is the size of electric current that must flow.