MCQ
Percentage of silver in German silver is
- A$2\%$
- B$1\%$
- C$5\%$
- ✓$0\%$
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$Zn \to Z{n^{2 + }} + 2{e^ - };{E^o} = 0.76\,\,V$,$Fe \to F{e^{2 + }} + 2{e^ - };{E^o} = 0.44\,\,V$ what will be the emf of cell, whose cell-reaction is ............ $\mathrm{V}$
$F{e^{2 + }}(aq) + Zn \to Z{n^{2 + }}(aq) + Fe$
(Atomic numbers $Ce =58, Gd =64$ and $Eu =63 .)$
$(a)$ $\left( NH _{4}\right)_{2}\left[ Ce \left( NO _{3}\right)_{6}\right]$
$(b)$ $Gd \left( NO _{3}\right)_{3}$ and
$(c)$ $Eu \left( NO _{3}\right)_{3}$
Answer is :