An alloy of $$Pb-Ag$$ weighing $$1.08$$ g was dissolved in dilute $$HNO3$$ and the volume made to $$100$$ mL. A silver electrode was dipped in the solution and EMF of the cell set up was $$Pt(s)$$,$$H2(g)H+(1M)$$∥$$Ag(aq$$.$$)+Ag(s)0.62$$ V. The percentage of Ag in the alloy is: Ecell ∘$$=0.80V$$,$$2.303RT/F=0.06$$ at $$25$$∘C

Question

An alloy of $$Pb-Ag$$ weighing $$1.08$$ g was dissolved in dilute $$HNO3$$ and the volume made to $$100$$ mL. A silver electrode was dipped in the solution and EMF of the cell set up was $$Pt(s)$$,$$H2(g)H+(1M)$$∥$$Ag(aq$$.$$)+Ag(s)0.62$$ V. The percentage of Ag in the alloy is: Ecell ∘$$=0.80V$$,$$2.303RT/F=0.06$$ at $$25$$∘C

Infinity Learn · Accepted Answer

$$H2+2Ag+$$⟶$$2H++2AgEcell$$ $$=Ecell$$ ∘−$$2.303RTnFlog$$⁡$$1Ag+20.62=0.80+0.06log$$⁡$$Ag+or$$ log⁡$$Ag+=$$−$$0.180.06=$$−$$3Ag+=1.0$$×$$10$$−$$3M=1.0$$×$$10$$−$$3$$×$$108=0.108gL$$−$$1$$∴ Amount of Ag in $$100$$ mL solution $$=$$ $$0.0108$$ g∴ % $$Ag=0.01081.08$$×$$100=1$$

Electrochemical cells

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Electrochemical cells

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An alloy of Pb-Ag weighing 1.08 g was dissolved in dilute HNO3 and the volume made to 100 mL. A silver electrode was dipped in the solution and EMF of the cell set up was Pt(s),H2(g)H+(1M)∥Ag(aq.)+Ag(s)0.62 V. The percentage of Ag in the alloy is: Ecell ∘=0.80V,2.303RT/F=0.06 at 25∘C

H2+2Ag+⟶2H++2AgEcell =Ecell ∘−2.303RTnFlog⁡1Ag+20.62=0.80+0.06log⁡Ag+or log⁡Ag+=−0.180.06=−3Ag+=1.0×10−3M=1.0×10−3×108=0.108gL−1∴ Amount of Ag in 100 mL solution = 0.0108 g∴ % Ag=0.01081.08×100=1

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