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 ...

Answer Details

The two voltameters are in the same circuit in series, so the same current flows through both for the same length of time. That means the quantity of charge \(Q = It\) passed through each is identical, and this shared value of \(Q\) is the bridge between the two metals.

By Faraday's first law, \(m = ZQ\), so for each metal \(Q = m/Z\). Equating the charges,

\[\frac{m_{\text{Ag}}}{Z_{\text{Ag}}} = \frac{m_{\text{Cu}}}{Z_{\text{Cu}}} \quad\Rightarrow\quad \frac{Z_{\text{Ag}}}{Z_{\text{Cu}}} = \frac{m_{\text{Ag}}}{m_{\text{Cu}}}.\]

Substituting the masses and the known electrochemical equivalent of copper,

\[Z_{\text{Ag}} = Z_{\text{Cu}}\times \frac{m_{\text{Ag}}}{m_{\text{Cu}}} = 0.00028\times \frac{1.2}{0.2} = 0.00028\times 6 = 1.68\times 10^{-3}\ \text{g C}^{-1}.\]

Notice that neither the current nor the time was needed, and neither was given: because the charge is common to both cells, it cancels out of the ratio. Recognising that cancellation is the real skill being tested here.

The likely error is inverting the mass ratio, using \(0.2/1.2\), which would give a value smaller than the copper figure. Check the sense of your answer physically: silver has a much larger mass deposited for the same charge, so its electrochemical equivalent, the mass per coulomb, must be the larger of the two. That is consistent with the chemistry, since each silver ion \(\text{Ag}^{+}\) carries only one elementary charge while each \(\text{Cu}^{2+}\) ion carries two.