One half cell in a voltaic cell is constructed by dipping silver rod in $$\text{AgNO}_3$$ solution of unknown concentration, other half cell is Zn rod dipped in 1 molar solution of $$\text{ZnSO}_4$$. A voltage of $$1.60\,\text{V}$$ is measured at $$298\,\text{K}$$ for this cell. What is the concentration of $$\text{Ag}^+$$ ions used in terms of $$\log x$$ $$(x = [\text{Ag}^+])$$?

$$E^\ominus_{\text{Zn}^{2+}/\text{Zn}} = -0.76\,\text{V}, \quad$$ $$E^\ominus_{\text{Ag}^{+}/\text{Ag}} = +0.80\,\text{V}, \quad$$ $$\frac{2.303RT}{F} = 0.059\,\text{V}$$

To determine the concentration of (Ag^+) ions, we use the Nernst equation, which relates the cell potential of an electrochemical cell to the concentrations of the reacting species under non-standard conditions.

The overall cell reaction is

$$Zn(s)+2Ag^+(aq)\rightarrow Zn^{2+}(aq)+2Ag(s).$$

The standard cell potential is calculated as

$$E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}=0.80-(-0.76)=1.56\text{ V}.$$

Applying the Nernst equation,

$$E_{\text{cell}}=E^\circ_{\text{cell}}-\frac{0.059}{n}\log Q,$$

where (n=2) and

$$Q=\frac{[Zn^{2+}]}{[Ag^+]^2}.$$

Substituting the given values (E_{\text{cell}}=1.60\text{ V}), (E^\circ_{\text{cell}}=1.56\text{ V}), ([Zn^{2+}]=1\text{ M}), and letting ([Ag^+]=x),

$$1.60=1.56-\frac{0.059}{2}\log\left(\frac{1}{x^2}\right).$$

Rearranging,

$$0.04=-0.0295\log\left(\frac{1}{x^2}\right),$$

$$0.04=0.059\log x.$$

Hence,

$$\log x=\frac{0.04}{0.059}=\frac{4}{5.9}.$$

Therefore, the correct choice is (\frac{4}{5.9}), which corresponds to option B.