At 298K, the EMF of the cell: Mg(s) |Mg2+(Q)||Ag+(0.01)|Ag(s)is 3.022V. Calculate the value ‘Q’.(Given: EoMg2+/Mg = -2.37V and EoAg+/Ag = 0.80V)

To solve this problem, we need to use the Nernst equation which is:

E = E° - (RT/nF) * lnQ

Where: E is the cell potential, E° is the standard cell potential, R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, n is the number of moles of electrons transferred in the balanced redox reaction, F is the Faraday constant (96485 C/mol), and Q is the reaction quotient.

First, we need to calculate the standard cell potential (E°). This is the difference between the reduction potentials of the two half-cells.

E° = E°(cathode) - E°(anode) E° = E°Ag+/Ag - E°Mg2+/Mg E° = 0.80V - (-2.37V) E° = 3.17V

Next, we substitute the known values into the Nernst equation and solve for Q:

3.022V = 3.17V - (8.314 J/(mol·K) * 298K / (2 * 96485 C/mol)) * lnQ -0.148V = -(8.314 J/(mol·K) * 298K / (2 * 96485 C/mol)) * lnQ 0.148V = (8.314 J/(mol·K) * 298K / (2 * 96485 C/mol)) * lnQ

Solving for lnQ gives:

lnQ = 0.148V / (8.314 J/(mol·K) * 298K / (2 * 96485 C/mol)) lnQ = 0.148V * (2 * 96485 C/mol) / (8.314 J/(mol·K) * 298K) lnQ = 0.148V * 192970 C/mol / 2474.652 J/mol lnQ = 11.56

Finally, we take the exponential of both sides to solve for Q:

Q = e^11.56 Q = 105,000

So, the value of Q is approximately 105,000.