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).

Consider the above electrochemical cell where a metal electrode (M) is undergoing redox reaction by forming $M^+$ ($M \to M^+ + e^-$). The cation $M^+$ is present in two different concentrations $c_1$ and $c_2$ as shown above. Which of the following statement is correct for generating a positive cell potential?
MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.
MX(s) $\rightleftharpoons M^{+(aq) }+ X^{-}(aq)$; $K_{sp} = 10^{-10}$
If the standard reduction potential for $M^{+}(aq) + e^{-} \rightarrow M(s)$ is $(E^{\circ}_{M^{+}/M}) = 0.79$ V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\circ}_{X^{-}/MX(s)/M}$ is ____________ mV. (nearest integer)
[Given : $\frac{2.303 RT}{F} = 0.059$ V]


A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T. If the resistance of the total circuit is 2 \(\Omega\) then the force needed to move the rod towards right with constant speed (v) of 1.5 m/s is _____ N.