Question

The specific conductance of a solution is 0.2 $- 2.36V.$ and conductivity is 0.04 $Mg^{2 +}|Mg$. The cell constant would be.

• A. 1 $- 2.36V$
• B. 0 $Fe_{(s)}|Fe^{2 +}(0.001M)||H^{+}(1M)|H_{2(g)}(1bar)|Pt_{(s)}$
• C. 5 $E_{cell} = Eº_{cell} - \frac{0.591}{2}\log\frac{\lbrack Fe^{2 +}\rbrack\lbrack H^{+}\rbrack^{2}}{\lbrack Fe\rbrack\lbrack H_{2}\rbrack}$
• D. 0.2 $E_{cell} = Eº_{cell} - \frac{0.591}{2}\log\frac{\lbrack Fe\rbrack\lbrack H^{+}\rbrack^{2}}{\lbrack Fe^{2 +}\rbrack\lbrack H_{2}\rbrack}$

Answer 1

Answer: (C)

5 $E_{cell} = Eº_{cell} - \frac{0.591}{2}\log\frac{\lbrack Fe^{2 +}\rbrack\lbrack H^{+}\rbrack^{2}}{\lbrack Fe\rbrack\lbrack H_{2}\rbrack}$

Answer 2

$K = C \times \text{Cellconstant} = \frac{K}{C} = \frac{0.2}{0.04} = 5cm^{- 1}$.