Excess of a solid metal M is added to an aqueous solution containing 0.1 M of N²⁺ ions. The system is allowed to reach equilibrium at 298 K according to the reaction: M(s) + N²⁺(aq) M²⁺(aq) + N(s) The standard reduction potentials are E^ _ M²⁺/M = -0.40 V and E^ _ N²⁺/N = 0.19 V . If the equilibrium concentration of N²⁺ ions is 10^ -x M , the value of x is ________. (Given: 2.303RT F = 0.059 V )

Step-by-step solution

First, calculate the standard cell potential E^ _ cell for the reaction: E^ _ cell = E^ _ cathode - E^ _ anode Here, N²⁺ is reduced (cathode) and M is oxidized (anode). E^ _ cell = 0.19 - (-0.40) = 0.59 V The equilibrium constant K is related to E^ _ cell by: E^ _ cell = 2.303RT nF K For this reaction, n = 2 . 0.59 = 0.059 2 K K = 0.59 2 0.059 = 20 K = 10²⁰ Since K is extremely large, the reaction proceeds almost to completion. The initial concentration of N²⁺ is 0.1 M . At equilibrium, almost all N²⁺ has reacted,