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: \( m_{\mathrm{Cu}} = 0.2 \, \mathrm{g} \), \( m_{\mathrm{Ag}} = 1.2 \, \mathrm{g} \), \( Z_{\mathrm{Cu}} = 0.00028 \, \mathrm{g \, C^{-1}} \) \(Z_{\mathrm{Ag}} = \frac{1.2 \times 0.00028}{0.2} = \frac{0.000336}{0.2} = 0.00168 \, \mathrm{g \, C^{-1}} \)