Question

An electrochemical cell has a constant potential of 2 V and provides a steady current of 1 A. How many electrons move from the anode to the cathode in 1000 seconds? (Use F = 1x105 C/mol e- )

Answer 1

Answer:

$n=6.25\times 10^{21}$

Explanation:

Potential across the electrochemical cell, V = 2 V

Current, I = 1 A

Time, t = 1000 s

Let n electrons move from the anode to the cathode. We know that electric current is calculated as :

$q=I\times t$

Also, q = ne

$ne=It$

$n=\dfrac{It}{e}$

$n=\dfrac{1\times 1000}{1.6\times 10^{-19}}$

$n=6.25\times 10^{21}$

So, $6.25\times 10^{21}$ electrons move from the anode to the cathode in 1000 seconds.