Question

Given $- nFEº$.

The equivalent conductance of the electrolytic cell is.

• A. $- E_{oxid}$
• B. $\rightarrow$
• C. $\rightarrow$
• D. $\rightarrow$

Answer 1

Answer: (A)

$- E_{oxid}$

Answer 2

$l/a = 0.5cm^{- 1},R = 50ohm$

$p = \frac{Ra}{l} = \frac{50}{0.5} = 100$

$\Lambda = k \times \frac{1000}{N} = \frac{1}{p} \times \frac{1000}{N} = \frac{1}{100} \times \frac{1000}{1}$

$10ohm^{- 1}cm^{2}gmeq^{- 1}$