Question

$\Lambda_m$ of a decimolar $CaCl_2$ solution is $100 \frac{S cm^2}{mol}$ at 298 K. Cell with electrodes 1.5 $cm^2$ in area and $\frac{1}{2}$ cm apart is filled with decimolar $CaCl_2$. Current flown in amperes when the potential difference between electrodes is 5 volts is

Answer 1

0.15

Answer 2

To calculate the current flowing through the electrochemical cell, we follow these steps:

1. Calculate the Cell Constant ($G^*$): The cell constant is defined as the ratio of the distance between the electrodes ($l$) to the area of the electrodes ($A$). Given: $l = 0.5 \text{ cm}$, $A = 1.5 \text{ cm}^2$ $G^* = \frac{l}{A} = \frac{0.5 \text{ cm}}{1.5 \text{ cm}^2} = \frac{1}{3} \text{ cm}^{-1}$

2. Calculate the Specific Conductivity ($\kappa$): The molar conductivity ($\Lambda_m$) is related to the specific conductivity ($\kappa$) and the concentration ($C$) by the formula: $\Lambda_m = \frac{\kappa \times 1000}{C}$ Where $\Lambda_m$ is in S cm$^2$ mol$^{-1}$, $\kappa$ is in S cm$^{-1}$, and $C$ is in mol L$^{-1}$. Given: $\Lambda_m = 100 \text{ S cm}^2 \text{ mol}^{-1}$, $C = 0.1 \text{ M} = 0.1 \text{ mol L}^{-1}$ Rearranging the formula to find $\kappa$: $\kappa = \frac{\Lambda_m \times C}{1000}$ $\kappa = \frac{100 \text{ S cm}^2 \text{ mol}^{-1} \times 0.1 \text{ mol L}^{-1}}{1000}$ $\kappa = \frac{10}{1000} \text{ S cm}^{-1}$ $\kappa = 0.01 \text{ S cm}^{-1}$

3. Calculate the Conductance (G): Conductance ($G$) is related to specific conductivity ($\kappa$) and cell constant ($G^*$) by the formula: $G = \frac{\kappa}{G^*}$ $G = \frac{0.01 \text{ S cm}^{-1}}{1/3 \text{ cm}^{-1}}$ $G = 0.01 \times 3 \text{ S}$ $G = 0.03 \text{ S}$

4. Calculate the Current (I): According to Ohm's Law, the current ($I$) is given by the potential difference ($V$) divided by the resistance ($R$). Since conductance ($G$) is the reciprocal of resistance ($R = 1/G$), we can write: $I = \frac{V}{R} = V \times G$ Given: $V = 5 \text{ volts}$ $I = 5 \text{ V} \times 0.03 \text{ S}$ $I = 0.15 \text{ A}$

The current flown in amperes is 0.15 A.