Question

Suppose 870.mmol of electrons must be transported from one side of an electrochemical cell to another in 84. seconds. Calculate the size of electric current that must flow.

Answer 1

Answer:

The size of an electric current that must flow is 999.18 Ampere.

Explanation:

Moles of electrons passed, n = 870. mmol= 0.870 mol

1 mmol = 0.001 mol

Number of electrons =$N=n\times N_A$

Total charge on N electrons = Q = $1.602\times 10^{-19} C\times N$

Duration of time in which N electrons were passed = T = 84. seconds

Amount of current passes = I

$I=\frac{Q}{T}$

$I=\frac{1.602\times 10^{-19} C\times N}{84. s}$

$=\frac{1.602\times 10^{-19} C\times 0.870\times 6.022\times 10^{23}}{84. s}$

I = 999.18 Ampere

The size of an electric current that must flow is 999.18 Ampere.