From standard reduction potentials, calculate the equilibrium constant at 25 ∘c for the reaction 2mno−4(aq)+10cl−(aq)+16h+(aq)→2mn2+ (aq)+5cl2(g)+8h2o(l)

Explanation:

Step 1. Determine the cell potential

                                                                                      E°/V      

2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺+ 8H₂O                      1.507

10Cl⁻ ⟶ 5Cl₂ + 10e⁻                                                 -1.358 27

2MnO₄⁻ + 10Cl⁻ + 16H⁺⟶ 2Mn²⁺ + 5Cl₂ + 8H₂O      0.149

Step 2. Calculate K

The formula relating K and E is

E = 0.149 V

R = 8.314 J·K⁻¹mol⁻¹

T = 25 °C = 298.15 K

n = 10

F = 96 485 C/mol

The value of the equilibrium constant is

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2