电极反应方程式

电位-pH方程

a

2H⁺ + 2e = H₂

φ = −0.0592pH

b

O₂ + 4 H⁺ + 4e = 2H₂O

φ = 1.229 − 0.0592pH

c

SO2− 4 + 4H⁺ + 2e = SO₂ + 2H₂O

φ = −0.126 − 0.1182pH

1

Te + 2H⁺ + 2e = H₂Te

φ = −0.739 − 0.0592pH

2

HTe⁻ + H⁺=H₂Te

pH = 2.82

3

Te²⁻ + H⁺ = HTe⁻

pH = 11.0

4

Te + H⁺ + 2e = HTe⁻

φ = −0.8174-0.0296pH

5

Te + 2e = Te²⁻

φ = −1.143

6

Te⁴⁺ + 4e = Te

φ = 0.568

7

TeO₂ + 4H⁺ + 4e = Te + 2H₂O

φ = 0.521 − 0.0592pH

8

TeO2− 3 + 6 H⁺ + 4e = Te + 3H₂O

φ = 0.827 − 0.08875pH

9

TeO2− 3 + 2 H⁺ = TeO₂ + 2H₂O

pH = 10.36

10

TeO₂ + 4H⁺ = Te⁴⁺ + 2H₂O

pH = −0.79

11

H₂TeO₄ + 6H⁺ + 2e = Te⁴⁺ + 4H₂O

φ = 0.92 − 0.1775pH

12

H₂TeO₄ + 2H⁺ + 2e = TeO₂ + 2H₂O

φ = 1.020 − 0.0592pH

13

HTeO− 4 + 3H⁺ + 2e = TeO₂ + 2H₂O

φ = 1.202 − 0.08875pH

14

TeO2− 4 + 2H⁺ + 2e = TeO2− 3 + H₂O

φ = 0.892 − 0.0592pH

15

TeO2− 4 + H⁺ = HTeO− 4

pH =10.49

16

HTeO− 4 + H⁺ = H₂TeO₄

pH = 5.62