Question

Suppose 650.mmol of electrons must be transported from one side of an electrochemical cell to another in 10.0 minutes. Calculate the size of electric current that must flow.

Answer 1

Answer:

The size of electric current must 104.5 Ampere.

Explanation:

Moles of electrons = 650 mmol = $650\times 0.001 mol$

mmol = 0.001 mol

1 mol =$N_A=6.022\times 10^{23}$ atoms/ions

Number of electrons = N

N = $650\times 0.001\times 6.022\times 10^{23}=3.9143\times 10^{23}$

Charge on an electron = $1.602\times 10^{-19} C$

Total charge on N electrons = Q

$Q=3.9143\times 10^{23}\times 1.602\times 10^{-19} C=62,707.086 C$

Duration of time = T = 10 min = 10 × 60 s

1 minute = 60 seconds

$Current(I)=\frac{Charge(Q)}{Time(T)}$

$I=\frac{62,707.086 C}{10\times 60 s}=104.5 A$

The size of electric current must 104.5 Ampere.