Copper of 0.2g and silver of 1.2g are deposited when current is passed through copper and silver voltameter. Calculate the electrochemical equivalent, Z of silver if that of copper is 0.00028gC\(^{-1}\)
From Faraday's first law, the mass deposited \( m = Z Q \), where \( Z \) is the electrochemical equivalent and \( Q \) is the charge passed.
Since the same charge \( Q \) passes through both voltmeters:
\(Q = \frac{m_{\mathrm{Cu}}}{Z_{\mathrm{Cu}}} = \frac{m_{\mathrm{Ag}}}{Z_{\mathrm{Ag}}} \)
Rearranging for \( Z_{\mathrm{Ag}} \):
\(Z_{\mathrm{Ag}} = \frac{m_{\mathrm{Ag}} \cdot Z_{\mathrm{Cu}}}{m_{\mathrm{Cu}}} \)
But: \( m_{\mathrm{Cu}} = 0.2 \, \mathrm{g} \), \( m_{\mathrm{Ag}} = 1.2 \, \mathrm{g} \), \( Z_{\mathrm{Cu}} = 0.00028 \, \mathrm{g \, C^{-1}} \)
\(Z_{\mathrm{Ag}} = \frac{1.2 \times 0.00028}{0.2} = \frac{0.000336}{0.2} = 0.00168 \, \mathrm{g \, C^{-1}} \)
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