# 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

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

For the metal ﻿$\Rightarrow$﻿ Oxidation ﻿$=$﻿ Anode

In order to produce ﻿$H_{2}$﻿ gas, ﻿$H^{-}$﻿ 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 ﻿$H_{2}$﻿ gas. (H can never work as an Anode)

Anything above H can't produce H gas.

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

Aqua Vegia ﻿$HCI = HNO_{3}$﻿ - 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 ﻿$\Delta G^{\circ }$﻿

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

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

Spontaneous

﻿$\Delta G^{\circ }<1$﻿ Negative

﻿$E^{\circ }>1$﻿ Positive

﻿$K>1$﻿

At Equilibrium

﻿$\Delta G^{\circ }=O$﻿

﻿$K=1$﻿

﻿$E^{\circ }_{cell}=O$﻿

﻿$E_{cell}=E^{\circ }_{cell}-\frac{0.1592v}{n}logQ$﻿ (Relationship between ﻿$E_{cell}$﻿ and ﻿$E^{\circ }_{cell}$﻿ )

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

﻿$E^{\circ }_{cell}=E_{right}-E_{left}$﻿