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Question

An alloy of PbAg weighing 1.08g was dissolved in dilute HNO3 and the volume made to 100 mL. A silver electrode was dipped in the solution and the cell is set up as
Pt(s),H2(g)|H+(1M)||Ag+(aq)|Ag(s)
The emf of the cell was 0.62 V. If EoCell=0.80V. What is the percentage of Ag in the alloy?[At 25oC, 2.303 RT/F=0.06]

A
25
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B
2.50
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C
10
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D
1
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Solution

The correct option is D 1
cell reaction is : H2+2Ag+2H++2Ag .
Ec11=E0ce112303RT2Flog1[Ag+]2 (Nernst's Equation)

0.62=0.80+0.06log[Ag+]log[Ag+]=0.180.06=3.[Ag+]=103 m.

0r, [Ag+]=(103×108)gm/L=0.108gm/L.

Amount of Ag in 100 mL solution =0.10810gm=0.0108.

Thus, 1. of Ag in alloy =0.01081.08×100=1%
so, option (D) is correct.

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