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For H^(+) , lambda^(0) = 349.8 S cm^(2) ...

For `H^(+) , lambda^(0) = 349.8 S cm^(2) mol^(-1)` . Calculate the velocity of `H^(+)` electrodes are 5 cm apart to which 1 volt of potential is applied. If the answer is `7.0 xx 10^(-x) ` cm/s then x is -

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The correct Answer is:
4
Velocity of `H^(oplus)`=(Ionic mobility ` xx ` potential gradient )
` = ((^^_M^(0))/(F)) xx (("voltage")/("distance")) = (349.8 S - cm^(2) - " mole"^(-1))/(96500) xx (1" volt")/(5 cm ) = 7.249 xx 10^(-4) cm//sec`
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