12) If the E_(equ) value of galvanic cell below is 1.50 V. What should be the concentration of [Ag'] Zn(s)vert Zn^+2(0.018M)Vert Ag^+(ldots M)vert Ag(s) (E^0Zn^ast /n.a^m-0.763V;E^circ Ag'/A_(E)=0.799V) A) 0.009M B) 0.012 M C) 0.016 m D) 0.024 M E) 0.083 M

To find the concentration of $[Ag']$ in the given galvanic cell, we can use the Nernst equation. The Nernst equation relates the cell potential to the concentrations of the reactants and products.<br /><br />The Nernst equation is given by:<br /><br />$E_{cell} = E^0_{cell} - \frac{0.0592}{n} \log Q$<br /><br />where $E_{cell}$ is the cell potential, $E^0_{cell}$ is the standard cell potential, $n$ is the number of moles of electrons transferred, and $Q$ is the reaction quotient.<br /><br />In this case, the standard cell potential ($E^0_{cell}$) can be calculated as:<br /><br />$E^0_{cell} = E^0_{cathode} - E^0_{anode} = 0.799V - (-0.763V) = 1.562V$<br /><br />The reaction quotient ($Q$) is given by:<br /><br />$Q = \frac{[Ag^+]}{[Zn^{2+}]}$<br /><br />We are given that $E_{equ} = 1.50V$ and $[Zn^{2+}] = 0.018M$. We need to find $[Ag^+]$.<br /><br />Substituting the values into the Nernst equation:<br /><br />$1.50V = 1.562V - \frac{0.0592}{2} \log \frac{[Ag^+]}{0.018M}$<br /><br />Solving for $[Ag^+]$, we get:<br /><br />$[Ag^+] = 0.024M$<br /><br />Therefore, the correct answer is D) 0.024 M.