MCQ
The relation between Faraday constant $(F)$, chemical equivalent $(E)$ and electrochemical equivalent $(Z)$ is
- A$F = EZ$
- B$F = \frac{Z}{E}$
- ✓$F = \frac{E}{Z}$
- D$F = \frac{E}{Z^2}$
where $E$ is chemical equivalent, $Z$ is electrochemical equivalent and $F$ is faraday constant.
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The $E _{\text {ceil }}$ for the given cell is $0.1115\,V$ at $298\,K$ when $\frac{\left[ M ^{+}( aq )\right]}{\left[ M ^{3+}( aq )\right]}=10^{ a }$
The value of a is
Given : $E _{ M ^{3+} / M ^{+}}=0.2\,V$
$\frac{2.303 RT }{ F }=0.059\,V$