JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
\(250\ mL\) of a waste solution obtained from the workshop of a goldsmith contains \(0.1\, M AgNO _{3}\) and \(0.1\, M\) AuCl. The solution was electrolyzed at \(2\, V\) by passing a current of \(1 \,A\) for \(15\) minutes. The metal/metals electropositive will be \(\left(E_{A g^{+} / A g}^{0}=0.80\, V, E_{A n^{+} / A u}^{0}=1.69\, V\right)\)
- A only silver
- B only gold
- C silver and gold in equal mass proportion
- D silver and gold in proportion to their atomic weights
Answer & Solution
Correct Answer
(D) silver and gold in proportion to their atomic weights
Step-by-step Solution
Detailed explanation
As voltage is \({ }^{\prime} 2 V ^{\prime}\) so both \(Ag ^{+} \& Au ^{+}\) will reduce and their equal gm equivalent will reduce so \(gmeq\, Ag = gmeq\) of \(Au\) \(\frac{ Wt _{ Ag }}{ E _{ qw t _{ Ag }}}=\frac{ Wt _{ Au }}{ E _{ qwt_{ Au}} }\) So…
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