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.

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 =

Total charge on N electrons = Q =

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

Amount of current passes = I

I = 999.18 Ampere

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

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