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For H^(+) and Na^(+) the values of lambd...

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

Text Solution

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The correct Answer is:
`mu^(@)H^(+) = 3.62 xx 10^(-3) cm^(2) volt^(-1)sec^(-1), mu_(Na)^(@) = 5.20 xx 10^(-14)cm^(2) volt^(-1) sec^(-1)`
We have
`u_(H^(+))^(@)=(lamda_(H^(+))^(@))/(F)=(349.8)/(96500)=3.62 xx 10^(-3) 10^(-3) cm^(2) "volt"^(-1) s^(-1)`
`u_(Na^(+))^(@)=(lamda_(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)")`
`:.` velocity of `H^(+)=3.62 xx10^(-3) xx(2)/(5) =1.45 xx 10^(-3) "cms"^(-1)`
velocity of `Na^(+)=5.20 xx 10^(-4) xx (2)/(5) =2.08 xx 10^(-4) "cms"^(-1)`.
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