A voltameter has a resistance of 2,Omega. It | vedclass

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(B) The mass $m$ deposited at the cathode is given by Faraday's law: $m = ZIt$,where $Z$ is the electrochemical equivalent,$I$ is the current through

Vedclass · Accepted Answer

$1.5$

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(B) The mass $m$ deposited at the cathode is given by Faraday's law: $m = ZIt$,where $Z$ is the electrochemical equivalent,$I$ is the current through the voltameter,and $t$ is time.Case $1$ (Series): The total resistance is $R_s = 2\,\Omega + 3\,\Omega = 5\,\Omega$. The current through the voltameter is $I_s = V / R_s = 10\,V / 5\,\Omega = 2\,A$. Given $m_s = 1\,g$.Case $2$ (Parallel): The voltameter $(2\,\Omega)$ and the resistor $(3\,\Omega)$ are connected in parallel across the $10\,V$ battery. The current through the voltameter is $I_p = V / R_v = 10\,V / 2\,\Omega = 5\,A$.Since $m \propto I$ for the same time $t$,we have $m_p / m_s = I_p / I_s$.$m_p = (5\,A / 2\,A) 	imes 1\,g = 2.5\,g$.The increase in mass is $\Delta m = m_p - m_s = 2.5\,g - 1\,g = 1.5\,g$.

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