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.)