A) 1 ampere
B) 1 coulomb
C) 1 faraday
D) None of the above
Correct Answer: C
Solution :
\[A{{g}^{+}}+{{e}^{-}}\to Ag;\,\,{{E}_{Ag}}=\frac{Atomic\,\,Mass}{1}=108\] Number of faraday \[=\frac{{{W}_{Ag}}}{{{E}_{Ag}}}=\frac{108}{108}=1\].
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