Two voltameters one of copper and another of silver, are joined in parallel. When a total charge $$q$$ flows through the voltameters, equal amount of metals are deposited. If the electrochemical equivalents of copper and silver are $$z_1$$ and $$z_2$$ respectively the charge which flows through the silver voltameter is

Let the total charge supplied by the source be $$q$$. Because the two voltameters are connected in parallel, this charge divides into

$$q_1$$ through the silver voltameter,
$$q_2$$ through the copper voltameter,

such that

$$q_1 + q_2 = q$$  $$-(1)$$

For an electrolytic cell the mass of metal deposited is given by Faraday’s law: $$m = z\,Q$$, where $$z$$ is the electro-chemical equivalent (E.C.E.) and $$Q$$ is the charge passed.

Hence, masses deposited in the two voltameters are

$$m_{\text{Ag}} = z_2\,q_1$$ (silver)
$$m_{\text{Cu}} = z_1\,q_2$$ (copper)

The problem states that the two masses are equal:

$$z_2\,q_1 = z_1\,q_2$$  $$-(2)$$

Solve equation $$-(2)$$ for one of the charges, say $$q_1$$:

$$q_1 = \frac{z_1}{z_2}\,q_2$$  $$-(3)$$

Substitute $$-(3)$$ into the charge-conservation relation $$-(1)$$:

$$\frac{z_1}{z_2}\,q_2 + q_2 = q$$

$$q_2\left(1 + \frac{z_1}{z_2}\right) = q$$

$$q_2 = \frac{q}{1 + \dfrac{z_1}{z_2}}$$  $$-(4)$$

The required charge is the one through the silver voltameter, i.e. $$q_1$$. Using $$-(3)$$ and $$-(4)$$:

$$q_1 = \frac{z_1}{z_2}\,\frac{q}{1 + \dfrac{z_1}{z_2}}$$

Multiply numerator and denominator by $$\dfrac{z_2}{z_1}$$ to simplify:

$$q_1 = \frac{q}{1 + \dfrac{z_2}{z_1}}$$

Therefore, the charge that flows through the silver voltameter is $$\displaystyle \boxed{\frac{q}{1 + \dfrac{z_2}{z_1}}}$$

Option B which is: $$\frac{q}{1 + \frac{z_2}{z_1}}$$