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An alloy of Pb-Ag weighing 1.08 g was di...

An alloy of Pb-Ag weighing 1.08 g was dissolved in dilute HNO, and the volume made to 100 mL. A silver electrode was dipped in the solution and the emf of the cell set-up `Pt(s), H_2(g) |H^(+)(1 M)||Ag^(+) (ag)|Ag(s)` was 0.62 V. If `E_("cell")^@` is 0.80 V, what is the percentage of Ag in the alloy? `(At 25^@C, RT//F =0.06)`

A

25

B

2.5

C

10

D

1

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The correct Answer is:
D
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An alloy of Pb-Ag weighing 1.08g was dissolved in dilute HNO_(3) and the volume made to 100 mL.A ? Silver electrode was dipped in the solution and the emf of the cell dipped in the solution and the emf of the cell set-up as Pt(s),H_(2)(g)|H^(+)(1M)||Ag^(+)(aq.)|Ag(s) was 0.62 V . If E_("cell")^(@) is 0.80 V , what is the percentage of Ag in the alloy ? (At 25^(@)C, RT//F=0.06 )

The measured voltage of the cell, Pt(s)|H_2(1.0"atm")|H^+(aq)||Ag^+(1.0M)|Ag(s) is 1.02 V " at " 25^@C . Given E_("cell")^@ is 0.80 V, calculate the pH of the solution .

An alloy of Pb-Ag weighing 54 mg was dissolved in desired amount of HNO_(3) & volume was made upto 500ml . An Ag electrode was dipped in solution and then connected to standard hydrogen electrode anode. Then calculate % of Ag in alloy. Given: E_(cell) = 0.5V, E_(Ag^(+)//Ag)^(@) = 0.8V (2.303RT)/(F) = 0.06

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