(1) io3- + 5i- + 6h+ ® 3i2 + 3h2o (2) i2 + 2s2o32- ® 2i- + s4o62-
Oxidation: Example: two electrons removed from iron: Fe --> Fe2+ + 2e- two electrons removed from iodide: 2I- --> I2 + 2e-
In the laboratory, copper(I) Iodide is prepared by simply mixing an aqueous solutions of potassium iodide and a soluble copper(II) salt such copper sulphate. : :: Cu2+ + 2I− → CuI2 The CuI2 immediately decomposes to iodine and insoluble copper(I) iodide, releasing I2. : :: 2 CuI2 → 2 CuI + I2
KI reacts with Cu2+ ions and then the CuI2 formed dicomposes to give insoluble CuI salt and I2. The iodine makes the solution brown. Cu2+ + 2I− → CuI2 2 CuI2 → 2 CuI + I2 Sodium thiosulfate can be added to this mixture. It reacts with the iodine giving a white ppt in a colourless solution.
The answer is NOT a)a) I(g) + e → I-(g)b) I2(g) → 2I(g)c) I(g) → I+(g) + ed) Na(g) + I(g) → NaI(s)e) Na(s) + 1/2I2(s) → NaI(s)
(2i)3 = 8i3 = 8(i2)(i) = 8(-1)(i) = -8i
x + y = -2 ∴ y = -2 - x xy = 2 ∴ x(-2 - x) = 2 ∴ -2x - x2 = 2 ∴ x2 + 2x + 2 = 0 ∴ x2 + 2x + 1 = -1 ∴ (x + 1)2 = -1 ∴ x + 1 = i ∴ x = i - 1 recall: y = - 2 - x ∴ y = - 2 - (i - 1) ∴ y = -2 - i + 1 ∴ y = - i - 1 ∴ x3 + y3 = (i - 1)3 + (- i - 1)3 = (i - 1)(i2 - 2i + 1) + (- i - 1)(i2 + 2i + 2) = i3 - 2i2 + i - i2 + 2i - 1 - i3 - 2i2 - 2i - i2 - 2i - 2 = i3 - i3 - 2i2 - i2 - 2i2 - i2 + i + 2i - 2i - 2i - 1 - 2 = -6i2 - i - 3 = 3 - i
Net ionic: Br2 + 2I- -> 2Br- + I2
It is 2I + 1. The coefficient is put before the variable.
2I- + Br2 ---> I2 + 2Br-
2KI+MnO_2+3H_2 SO_4→2I+2KHSO_4+MnSO_4+2H_2 O
Br2 + (2e)- --> 2 Br- 2I- --> I2 + (2e)-
4
Maybe, but not very stable because 2Fe3+ + 2I- --> 2Fe2+ + I2 is a favourable forward reaction (positive difference in (standard) electrode potential scale 73 (Fe3+) - 62 (I2) = 11 eV
Cl2-(g) + 2I-(aq) ---------> 2Cl-(aq) + I2-(g)
ClO- (aq) +2I- (aq) +2H ions (aq) ----> I2 (s) + Cl- (aq) + H20 (l)
Bromine is the oxidizing element.