For H+ the value of λ∞ =349.8 S cm2 eq-1. Calculate the mobilities of H+ ion and its velocity if it is in a cell in which the electrodes are 5 cm apart and to which a potential of 2 volts is applied.

Question

For H+ the value of λ∞ =349.8 S cm2 eq-1. Calculate the mobilities of H+ ion and its velocity if it is in a cell in which the electrodes are 5 cm apart and to which a potential of 2 volts is applied. (A) 3.575 × 10-3 cm2 volt-1 sec-1, 1.43 × 10-3 cm sec-1 (B) 5.2 × 10-4 cm2 volt-1 sec-1, 2.08 × 10-4 cm sec-1 (C) 3.2 × 10-4 cm2 volt-1 sec1, 1.05 × 10-3 cm sec-1 (D) 1.43 × 10-3 cm2 volt-1 sec-1, 3.575 × 10-3 cm sec-1

Tardigrade Admin · Accepted Answer

Velocity =(1/2×349.8)=1.48 ×10-3 cm/ sec Mobility =1.43 ×10-3×(5/2)=3.575×10-3

Q. For $H^{+}$ the value of $λ^{\infty} = 349.8 S c m^{2}$ $e q^{- 1}$ . Calculate the mobilities of $H^{+}$ ion and its velocity if it is in a cell in which the electrodes are $5$ cm apart and to which a potential of $2$ volts is applied.

$3.575 \times 1 0^{- 3} c m^{2} v o l t^{- 1} se c^{- 1}, 1.43 \times 1 0^{- 3} c m se c^{- 1}$

$5.2 \times 1 0^{- 4} c m^{2} v o l t^{- 1} se c^{- 1}, 2.08 \times 1 0^{- 4} c m se c^{- 1}$

$3.2 \times 1 0^{- 4} c m^{2} v o l t^{- 1} se c^{1}, 1.05 \times 1 0^{- 3} c m se c^{- 1}$

$1.43 \times 10 - 3 c m^{2} v o l t^{- 1} se c^{- 1}, 3.575 \times 1 0^{- 3} c m se c^{- 1}$

Velocity $= \frac{1}{2 \times 349.8} = 1.48 \times 1 0^{- 3} c m / sec$
Mobility $= 1.43 \times 1 0^{- 3} \times \frac{5}{2} = 3.575 \times 1 0^{- 3}$

Q. For H+ the value of λ∞=349.8Scm2 eq−1. Calculate the mobilities of H+ ion and its velocity if it is in a cell in which the electrodes are 5 cm apart and to which a potential of 2 volts is applied.

3.575×10−3cm2volt−1sec−1,1.43×10−3cmsec−1

5.2×10−4cm2volt−1sec−1,2.08×10−4cmsec−1

3.2×10−4cm2volt−1sec1,1.05×10−3cmsec−1

1.43×10−3cm2volt−1sec−1,3.575×10−3cmsec−1