The amount of a substance deposited during electrolysis is determined using Faraday’s laws of electrolysis. The formula is:
\[W = ZIt,\]
where:
- \(W\) is the mass of the substance deposited,
- \(Z\) is the electrochemical equivalent of the substance,
- \(I\) is the current passed, and
- \(t\) is the time for which the current is passed.
Step 1: Relating charge to electrochemical equivalent
We know that:
\[Q = It,\]
where \(Q\) is the total charge passed through the solution. Substituting this into the equation for \(W\), we get:
\[W = ZQ\]
Step 2: Deposition of silver
For one coulomb of charge (\(Q = 1 \, \text{C}\)), the mass of silver deposited is directly proportional to the electrochemical equivalent (\(Z\)) of silver. Thus:
\[W = ZQ = (\text{electrochemical equivalent of silver}).\]
Step 3: Conclusion
The quantity of silver deposited when one coulomb of charge is passed is equal to the electrochemical equivalent of silver. This matches the given option.
Final Answer: (4).
If the molar conductivity ($\Lambda_m$) of a 0.050 mol $L^{–1}$ solution of a monobasic weak acid is 90 S $cm^{2} mol^{–1}$, its extent (degree) of dissociation will be:
[Assume: $\Lambda^0$ = 349.6 S $cm^{2} mol^{–1}$ and $\Lambda^0_{\text{acid}}$ = 50.4 S$ cm^{2} mol^{–1}$]
Let \( S = \left\{ m \in \mathbb{Z} : A^m + A^m = 3I - A^{-6} \right\} \), where
\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \]Then \( n(S) \) is equal to ______.
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: