Consider the following redox reaction : MnO4- + H+ + H2C2O4 Mn2+ + H2O + CO2 The standard reduction potentials are given as below ( E0red ) : E MnO4-/Mn2+ = +1.51 V E CO2/H2C2O4 = -0.49 V If the equilibrium constant of the above reaction is given as Keq = 10x, then the value of x = (nearest integer)
Answer
338
Solution
Balanced: 2MnO4- + 6H+ + 5H2C2O4 2Mn2+ + 8H2O + 10CO2 ⇒ n = 10
E cell = E cathode - E anode = 1.51 − (−0.49) = 2.00 V
At equilibrium: K = n E cell0.0591
⇒ log K = 10 2.000.0591 = 338.4
∴ Keq = 10338 ⇒ x = 338
Answer: 338