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∴ % $$Ag=0.01081.08$$×$$100=1$$

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