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Valency and Oxidation State

If mA gram ...

Question

If $m_{A}$ gram of a metal $A$ displaces $m_{B}$ gram of another metal $B$ from its salt solution and if the equivalent mass are $E_{A}$ and $E_{B}$ respectively then equivalent mass of $A$ can be expressed as:

E_{A} = \frac{m_{A}}{m_{B}} \times E_{B}

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E_{A} = \frac{m_{A} \times m_{B}}{E_{B}}

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E_{A} = \frac{m_{B}}{m_{A}} \times E_{B}

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E_{A} = \sqrt{\frac{m_{A}}{m_{B}} \times E_{B}}

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Solution

The correct option is B $E_{A} = \frac{m_{A}}{m_{B}} \times E_{B}$
Given,
$M_{A}$ gram of metal $A$ displaces $M_{B}$ gram of metal $B$
Equivalent weight of $A = E_{A}$
Equivalent weight of $B = E_{B}$
As there is no change in amount of Electricity supplies for displacement,
According to Faraday's law of electrolysis:
The mass (m) of a substance liberated or dissolved by the same amount of electricity is directly proportional to their equivalent masses.
$\Rightarrow m \propto E$
$\frac{m_{A}}{m_{B}} = \frac{E_{A}}{E_{B}}$
$\Rightarrow E_{A} = \frac{m_{A}}{m_{B}} \times E_{B}$

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