Introductory Chemistry (5th Edition)

Published by Pearson
ISBN 10: 032191029X
ISBN 13: 978-0-32191-029-5

Chapter 16 - Oxidation and Reduction - Exercises - Problems - Page 605: 84

Answer

Cr, Pb & Zn - these metals will dissolve in HCl. Balanced redox reaction for each metal with HCl is as follows: $2Cr (s) + 6HCl (aq) -> 2CrCl_{3} (aq) + 3H_{2}(g) $ $Pb(s) + 2HCl(aq) -> PbCl_{2} (aq) + H_{2} (g)$ $Zn (s) + 2HCl (aq) -> ZnCl_{2}(aq) + H_{2}(g) $

Work Step by Step

Metal Activity series: $Li>K>Ca>Na>Mg>Al>Mn>Zn>Cr>Fe>Ni>Sn>Pb>H_{2}>Cu>Ag>Au$ Activity decreases from left to right. More active metal will be oxidized if the reaction occurs spontaneously. From metal activity series it is seen that Cr, Pb & Zn are more active than $H_{2}$. So, it can be concluded that these metals will be oxidized in presence of acid i.e. HCl. So, these two metals dissolve in HCl. On the other hand, Au is less active than $H_{2}$. So, it do not dissolve in HCl. Redox reaction of Cr with HCl: $Cr (s) + H^{+} (aq) -> Cr^{3+} + H_{2}(g) $ Oxidation reaction after balancing charge: $ Cr(s) -> Cr^{3+} + 3e^{-}$ Reduction reaction after balance charge and number of element on both side we get : $2H^{+} + 2e^{-} -> H_{2}$ Now, we have to combine both half reactions. But before combining we have to make sure that the electrons are eliminated from both reactions. So, we multiply oxidation reaction with co-efficient 2 and reduction reaction with co-efficient 3. Now, combining both equations we get: $2Cr(s) + 6H^{+}(aq) ->2CrCl_{3} (aq) + 3H_{2} (g)$ or, $2Cr (s) + 6HCl (aq) -> 2CrCl_{3} (aq) + 3H_{2} (g) $ Redox reaction of Pb with HCl: $Pb (s) + H^{+} (aq) -> Pb^{2+} + H_{2}(g) $ Oxidation reaction after balancing charge: $ Pb(s) -> Pb^{2+} + 2e^{-}$ Reduction reaction after balance charge and number of element on both side we get : $2H^{+} + 2e^{-} -> H_{2}$ Now, combining both half reactions we get: $Pb(s) + 2H^{+}(aq) -> PbCl_{2} (aq) + H_{2} (g)$ or, $Pb(s) + 2HCl(aq) -> PbCl_{2} (aq) + H_{2} (g)$ Redox reaction of Zn with HCl: $Zn (s) + H^{+} (aq) -> Zn^{2+} + H_{2}(g) $ Oxidation reaction after balancing charge: $ Zn(s) -> Zn^{2+} + 2e^{-}$ Reduction reaction after balance charge and number of element on both side we get : $2H^{+} + 2e^{-} -> H_{2}$ Now, combining both half reactions we get: $Zn(s) + 2H^{+}(aq) -> ZnCl_{2} (aq) + H_{2} (g)$ or, $Zn(s) + 2HCl(aq) -> ZnCl_{2} (aq) + H_{2} (g)$
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