Lecture 4: Why Some Metals Produce Hydrogen When Reacting With Acid

Why can some meTals produce H when reacting with an acid? Like Cu = Copper

2H=2eH2(g)2H^{-}=2e \rightarrow H_{2(g)} Reaction

For the metal \Rightarrow Oxidation == Anode

In order to produce H2H_{2} gas, HH^{-} has to be reduced. It means it has to be part of Cathode. So any metal below it in the series can work as Anode and hence when these metals are placed in acid produce H2H_{2} gas. (H can never work as an Anode)

Anything above H can't produce H gas.

Most metals dissolve in Nitric acid (HNO3)(HNO_{3})

Aqua Vegia HCI=HNO3HCI = HNO_{3} - Gold dissolves only in Aqua Vegia

Why do all metals dissolve in nitrate acid? Because nitrate is a strong oxidizing agent \rightarrow metal becomes positive ion \rightarrow Dissolves

To Find ΔG\Delta G^{\circ }

  1. Find EE^{\circ } of the cell = E Cathode - E Anode
  2. \Delta G^{\circ }=-RTlnK =-nFE^{\circ }_{cell}

Nernst Equation = Ecell=EcellRTnFlnQE_{cell}=E^{\circ }_{cell}-\frac{RT}{nF}lnQ


ΔG<1\Delta G^{\circ }<1 Negative

E>1E^{\circ }>1 Positive


At Equilibrium

ΔG=O\Delta G^{\circ }=O


Ecell=OE^{\circ }_{cell}=O

Ecell=Ecell0.1592vnlogQE_{cell}=E^{\circ }_{cell}-\frac{0.1592v}{n}logQ (Relationship between EcellE_{cell}  and EcellE^{\circ }_{cell} )

(RTF=0.0592v)(\frac{RT}{F}=0.0592v) when T=298.15KT=298.15K

Ecell=ErightEleftE^{\circ }_{cell}=E_{right}-E_{left}

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