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).


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]
The value of \[ \int_0^{2} \sqrt{\frac{x(x^2+x+1)}{(x+1)(x^4+x^2+1)}} \, dx \] is