Consider the following redox reaction: MnO 4 - + H + + H 2 C 2 O 4 ⇌ Mn 2 + + H 2 O + CO 2 The standard reduction potentials are given as below E red ° E 0 MnO 4 - / Mn 2 + = + 1 . 51 V ; E 0 CO 2 / H 2 C 2 O 4 = - 0 . 49 V If the equilibrium constant of the above reaction is given as K eq = 10 x , then the value of x = _______ (nearest integer)

Step-by-step solution

For the reaction at equilibrium, 2 MnO 4 - + 6 H + + 5 H 2 C 2 O 4 ⇌ 2 Mn 2 + + 8 H 2 O + 10 CO 2 . Given : E MnO 4 - / Mn 2 + 0 = 1 . 51 V and E 0 H 2 C 2 O 4 / CO 2 = 0 . 49 V . So, E cell 0 = 1 . 51 + 0 . 49 = 2 V . Number of electrons involved in reaction can be calculated as follows: Mn + 7 changes Mn 2 + so, each Mn gain five electrons. n = 10 Now, E cell 0 = 0 . 0591 n log K log K = 10 × 2 0 . 0591 log K = 338 . 4