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Fudgin [204]
3 years ago
13

For a particular redox reaction ClO2– is oxidized to ClO4– and Fe3 is reduced to Fe2 . Complete and balance the equation for thi

s reaction in basic solution. Phases are optional.
Please HELP, I don't know how to balance redox reactions in basic solutions...!
Chemistry
1 answer:
creativ13 [48]3 years ago
8 0
We have to balance the equation in basic medium:
ClO₂⁻ → ClO₄⁻
Chlorine atoms are balanced 
we will balance oxygen atoms by adding water to the side with less oxygen
2 H₂O + ClO₂⁻ → ClO₄⁻
Now balance hydrogens by adding H⁺ first
2 H₂O + ClO₂⁻ → 4 H⁺ + ClO₄⁻
The charge will be balanced by adding electrons to side where positive charge is more
2 H₂O + ClO₂⁻ →  4 e + 4 H⁺ + ClO₄⁻
Now balance H⁺ by adding same number of OH⁻ in both sides
2 H₂O + ClO₂⁻ + 4OH⁻ →  4 e + 4 H⁺ + 4OH⁻ + ClO₄⁻
H⁺ will neutralize OH⁻ to give water
2 H₂O + ClO₂⁻ + 4 OH⁻ → 4 e + 4 H₂O + ClO₄⁻
Balanced half reaction will be:
ClO₂⁻ + 4 OH⁻ → 4 e + 2 H₂O + ClO₄⁻ → (1)
Now balance the second half (reduction):
Fe⁺³ → Fe⁺²
Balance charge by adding electrons:
e + Fe⁺³ → Fe⁺² → (2) Multiply by 4 and add the two equations to get:
ClO₂⁻ + 4 OH⁻ + 4 Fe⁺³  → 4 Fe⁺² + 2 H₂O + ClO₄⁻
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calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w
kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is -22Jmole^-^1k^-^1

(b) The normal boiling point of water (J·K−1·mol−1) is -109Jmole^-^1K^-^1

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is  109J/mole

Explanation:

Lets calculate

(a) - General equation -

      (\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p = -5_m(\beta )+5_m(\alpha ) =  -\frac{\Delta H}{T}

 \alpha ,\beta → phases

ΔH → enthalpy of transition

T → temperature transition

 (\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p == -\frac{\Delta_fH}{T_f}

            = \frac{-6.008kJ/mole}{273.15K} ( \Delta_fH is the enthalpy of fusion of water)

           = -22Jmole^-^1k^-^1

(b) (\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}

                                  = \frac{40.656kJ/mole}{373.15K} (\Delta_v_a_p_o_u_rH is the enthalpy of vaporization)

                               = -109Jmole^-^1K^-^1

(c) \Delta\mu =\Delta\mu(l)-\Delta\mu(s) =-S_m\DeltaT

[\mu(l-5°C)-\mu(l,0°C)] =  [\mu(s-5°C)-\mu(s,0°C)]=-S_mΔT

\mu(l,-5°C)-\mu(s,-5°C)=-Sm\DeltaT [\mu(l,0

\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)

     = 109J/mole

6 0
2 years ago
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