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Question: The electrochemical equivalent of a metal is \(x\) \(gram\) \(columb{\,^{ - 1}}\). The equivalent we...

The electrochemical equivalent of a metal is xx gramgram columb −1columb{\,^{ - 1}}. The equivalent weight of metal is.
A) xx
B) x×96500x \times 96500
C) x96500\dfrac{x}{{96500}}
D) 1.6×10−19× x1.6 \times {10^{ - 19}} \times \,x

Explanation

Solution

We know that electrochemical equivalent of an element is defined as the mass of an element deposited at an electrode when one ampere of electricity is flows through an electrolytic cell for one second and it is mentioned asZZ.

Complete step by step answer:
We know that the electrochemical cell which performs a reaction during the introduction of electricity is implicit as a cell .Michael Faraday conducted an in intense examination on electrolysis of solutions and melts of electrolytes. He futured two laws to clarify the quantitative aspects of electrolysis generally referred to as Faraday’s laws of electrolysis i.e. first law of electrolysis and therefore the second law of electrolysis
We know Faraday’s first law of electrolysis. According to this Faraday’s first law of electrolysis the amount of an element accumulated in the electrode is relative to the amount of electric charge flows through the electrolyte through electrolysis.
W∝QW \propto Q
W=ZQW = ZQ
Where,
Q=quantity of electricity
Z is the proportionality constant,
Z=EFZ = \dfrac{E}{F}
Where F is the Faraday’s constant and the value of Faraday’s constant is 9650096500
E=xgcoloumb−1×96500coloumbE = xgcoloum{b^{ - 1}} \times 96500coloumb
E=x×96500gE = x \times 96500g.

Therefore, the option B is correct.

Note:
Now we can discuss about the Faraday’s – Second Law of Electrolysis:
During electrolysis, when an equivalent amount of electricity flows through the electrolytic solution, mixture of various substances liberate are relative to their chemical equivalent weights
From these laws of electrolysis, we will assume that the amount of electricity required for oxidation-reduction depends on the stoichiometry of the electrode reaction.
For example,
Na++e−→ NaN{a^ + } + {e^ - } \to {\text{ }}Na
One mole of the electron is required for the reduction of 1 mole of sodium ions. We all know that charge on one electron is adequate to1.6021×10−19C1.6021 \times {10^{-19}}C. Thus, the charge on one mole of electrons is equal to,
NA×1.6021×10−19C=6.02×1023×1.6021×10−19C=9.647×104CNA \times 1.6021 \times {10^{-19}}C = 6.02 \times {10^{23}} \times 1.6021 \times {10^{-19}}C = 9.647 \times {10^4}C.