For H^(+) and Na^(+) the values of lambda^(@) are 349.8 and 50.11 respectively. Calculate the mobilities of these ions and their velocities if they are in a cell in which the electrodes are 5 cm apart and to which a potential of 2 volt is applied.

For H^(+) and Na^(+) the values of lambda^(@) are 349.8 and 50.11 resp

1.	For H^(+) and Na^(+) the values of lambda^(@) are 349.8 and 50.11 respectively. Calculate the mobilities of these ions and their velocities if they are in a cell in which the electrodes are 5 cm apart and to which a potential of 2 volt is applied.
Answer» Solution :We have `U_(H^(+))^(@) = (lambda_(H^(+))^(@))/(F) = (349.8)/(96500)` `= 3.62 xx 10^(-3) CM^(2) "volt"^(-1)s^(-1)` `U_(Na^(+))^(@) = (lambda_(Na^(+))^(@))/(F) = (50.11)/(96500)` `= 5.20 xx 10^(-4) cm^(2) "volt"^(-1)s^(-1)` Further, we KNOW that `U^(@) = ("ionic velocity (cm/s)")/("pot. diff. (volt)/distance between the electrodes (cm)")` `THEREFORE` velocity of `H^(+) = 3.62 xx 10^(-3) xx (2)/(5)` `= 1.45 xx 10^(-3) cm s^(-1)` Velocity of `Na^(+) = 5.20 xx 10^(-4) xx (2)/(5)` `= 2.08 xx 10^(-4) cm s^(-1)`.