Question

i2+5cu^2+ +6h20 = 2io3^- +5cu+12h^+
what is the reducing agent
what is density of f2 gas at temp 273.15k and 0.500 ATM.

Answer 1

Ans 1

The given equation is

I2 + 5cu2+ + 6H2O = 2IO3- + 5Cu + 12H+

Reducing agent - the substance oxidizes during the reaction, or oxidation number increases, is reducing agent.

Determine the oxidation number of each substance

I in I2 = 0

Cu in Cu2+ = +2

H in H2O = +2

O in H2O = -2

I in IO3- = +5

O in IO3- = - 6

Cu in Cu = 0

H in H+ = +1

The two half reactions are

Oxidation reaction

I2 + 6H2O = 2IO3- + 12H+ + 10e-

I is being oxidized

I is reducing agent

Reduction reaction

Cu2+ + 2e- = Cu

Cu is being reduced

Cu is oxidizing agent

Ans 2

From the ideal gas equation

Pressure x volume = moles x Gas constant x Temperature

Pressure x Molecular weight =( mass/volume) x Gas constant x Temperature

0.500 atm x 38 g/mol = density x 0.0821 L-atm/mol-K x 273.15K

19 = density x 22.425615

Density = 0.847 g/L x 1L/1000 mL

= 8.47 x 10^-3 g/mL