Question

The potential difference across the terminals of a battery of emf 12 V and internal resistance 2 Ω drops to 10 V when it is connected to a silver voltameter. Find the silver deposited at the cathode in half an hour. Atomic weight of silver is 107.9 g mol⁻¹.

Answer 1

Let i be the current through the circuit.
Emf of battery, E = 12 V
Voltage drop across the voltameter, V = 10 V
Internal resistance of the battery, r = 2 Ω
Applying Kirchoff's Law in the circuit, we get:
$$\begin{gathered} E=V+ir \\ \Rightarrow i=\frac{E-V}{r}=\frac{12-10}{2}=1\mathrm{A} \end{gathered}$$
Using the formula m = Zit, we get:
$m=\frac{107.9}{96500}\times 1\times 0.5\times 3600=2.01\mathrm{g}$