Question

Given the following standard reduction potentials

Ag⁺(aq) + e^- ↔ Ag(s)                         E = 0.80 V
AgCN(s) + e^- ↔ Ag(s) + CN-(aq)         E = -0.01 V

Calculate the solubility product of AgCN at 25°C.

		4.3 &αχυτε; 10−14
		2.3 &αχυτε; 1013
		2.0 &αχυτε; 10−14
		5.1 &αχυτε; 1013
		None of the above

Answer 1

Step 1: Firstly we have to find E⁰_(cell) for the reaction

For this, reverse the first equation and add them, as follows

Ag(s) <--> Ag⁺(aq) + e-      E⁰ = -0.80 V

AgCN (s) + e^- <----> Ag(s) + CN^-(aq)    E⁰ = -0.01 V

AgCN(s) <--->g+(aq) + CN-(aq)   E⁰_Cell = (-0.80 +(-0.01)) V = -0.81 V

*Note: When reversing any half-cell equation, the sign of E0 also gets reversed.

Step 2: Use the Nernst Equation:

E_cell = E° - (0.0591 / n) log K; where n is the number of electrons lost or gained and K is the equilibrium constant

Now remembering that when equilibrium is reached at saturated solution (the point when K=K_sp), the E_cell becomes zero

0 = -0.81 - (0.0591 / 1) log K

=> 0.81 / -0.0591 = log K

=> log K = -13.71

=> K = 1.95 × 10^-14

^{In this case it is K_sp, as we had taken E_cell to be zero}

^{So assuming the options given to be approximately true and in scientific notation, The third option from the top (2.0} × 10^-14) is the correct answer.