Two voltameters,one of copper and another of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) According to Faraday's law of electrolysis,the mass $m$ deposited is given by $m = zq$,where $z$ is the electrochemical equivalent and $q$ is the

Vedclass · Accepted Answer

$\frac{q}{1 + \frac{z_2}{z_1}}$

Vedclass · Answer

(D) According to Faraday's law of electrolysis,the mass $m$ deposited is given by $m = zq$,where $z$ is the electrochemical equivalent and $q$ is the charge.Since the masses deposited are equal,we have $m_1 = m_2$,which implies $z_1 q_1 = z_2 q_2$.From this,the ratio of charges is $\frac{q_1}{q_2} = \frac{z_2}{z_1}$.We know that the total charge $q = q_1 + q_2$.We want to find the charge $q_2$ flowing through the silver voltameter.From the ratio,$q_1 = q_2 \frac{z_2}{z_1}$.Substituting this into the total charge equation: $q = q_2 \frac{z_2}{z_1} + q_2 = q_2 (1 + \frac{z_2}{z_1})$.Solving for $q_2$,we get $q_2 = \frac{q}{1 + \frac{z_2}{z_1}}$.

Similar Questions

$A$ heater is rated $100\ W, 200\ V$. It is cut into two equal parts. Both parts are connected in parallel to the same $200\ V$ source. The energy released per second in this new combination is ............. $W$.

In the circuit shown below,the cell has an $e.m.f.$ of $10\,V$ and internal resistance of $1\,\Omega$. The other resistances are shown in the figure. The potential difference ${V_A} - {V_B}$ is ................ $V$.

$A$ current of $9\text{A}$ enters point $P$ of an equilateral triangle $PQR$ having three wires of $3\Omega$ each and leaves by point $R$. The currents $I_1$ and $I_2$ are respectively

Refer to the figure given below, current between terminals $A$ and $B$ is . . . . . . $A$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module