Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy ΔG0 for the following redox reaction. Round your answer to 3 significant digits. 10Cl^(−)(aq) + 2MnO4^(−)(aq) + 16H^(+)(aq) = 5Cl2(g) + 2Mn^(+2)(aq) + 8H2O(l)

The standard reaction free energy = -144 kj

Explanation:

According to Electrochemical series

Standard reduction potential of

Cl₂  + 2 e⁻  → 2 Cl⁻                                 E⁰ = +1.36 volt (i)MnO₄⁻ + 8H⁺ + 5 e⁻ → Mn²⁺  + 4 H₂O   E⁰ = +1.51 volt (ii)

Since Permanganate have more positive reduction potential so it is used as a cathode half cell and chlorine as a anode half cell

Cathode half cell (Reduction)     MnO₄⁻ + 8H⁺ + 5 e⁻ → Mn²⁺  + 4 H₂O ...(3)

Anode half cell    (Oxidation)       2 Cl⁻    →   Cl₂  + 2 e⁻ (4)

Multiplication by 2 of equation (3) and by 5 of equation(4)

10Cl⁻(aq) + 2MnO₄⁻(aq) + 16H⁺(aq) = 5Cl₂(g) + 2Mn²⁺(aq) + 8H₂O(l)

E.M.F of the combined cell reaction

                 E⁰ = E⁰cathode - E⁰anode = 1.51 - 1.36 = 0.15 v

               ⇒ ΔG⁰ reaction = - nF E⁰cell

n = no, of electron exchange = 10

              ⇒ΔG⁰ reaction = - 10 x 96500 x 0.15

                                        = - 144750 j

                                        = - 144 kj (round off by 3 significant fig.)

Cr(oh)3 -> cr(3+) + 3oh-

True is ur answer : )