Question

Silver and copper voltameters are connected in parallel with a battery of e.m.f. 12 V. In 30 minute 1 g of silver and 1.8 g of copper are liberated. The energy supplied by the battery is
$[Z_{Ag}=11.2\times {10}^{-4}g{c}^{-1};Z_{Cu}=6.6\times {10}^{-4}g{c}^{-1}]$

• A. 720 J
• B. 2.41 J
• C. 24.12 J
• D. $4.34\times {10}^{4}J$

Answer 1

Answer: (D)

$4.34\times {10}^{4}J$

Using Faraday's law of electrolysis:

$m_{Cu}=Z_{Cu}{I}_{Cu}t$ and $m_{Ag}=Z_{Ag}{I}_{Ag}t$

As voltameters are connected in parallel, so, $I=I_{Cu}+I_{Ag}$

Energy $=Pt=VIt=Vt[I_{Cu}+I_{Ag}]=V[\frac{m_{Cu}}{Z_{Cu}}+\frac{m_{Ag}}{Z_{Ag}}]=12[\frac{1.8}{6.6{\times }^{-4}}+\frac{1}{11.2\times {10}^{-4}}]=12\times {10}^4\left[0.362\right]=4.34\times {10}^{4}J$