Emf of the following cell at 298 K in V is $$x \times 10^{-2}$$
$$Zn|Zn^{2+}(0.1M)||Ag^+(0.01M)|Ag$$
The value of $$x$$ is ______ (Rounded off to the nearest integer)
[Given: $$E^\theta_{Zn^{2+}/Zn} = -0.76$$ V; $$E^\theta_{Ag^+/Ag} = +0.80$$ V; $$\frac{2.303RT}{F} = 0.059$$]

The cell is $$Zn|Zn^{2+}(0.1\text{ M})||Ag^+(0.01\text{ M})|Ag$$. The cell reaction is: $$Zn + 2Ag^+ \rightarrow Zn^{2+} + 2Ag$$.

The standard cell EMF is $$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.80 - (-0.76) = 1.56$$ V.

Using the Nernst equation: $$E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Ag^+]^2}$$.

Here $$n = 2$$ (two electrons transferred), $$[Zn^{2+}] = 0.1$$ M, and $$[Ag^+] = 0.01$$ M.

The reaction quotient is $$Q = \frac{[Zn^{2+}]}{[Ag^+]^2} = \frac{0.1}{(0.01)^2} = \frac{0.1}{0.0001} = 1000$$.

Therefore: $$E_{cell} = 1.56 - \frac{0.059}{2} \log(1000) = 1.56 - \frac{0.059}{2} \times 3 = 1.56 - 0.0885 = 1.4715$$ V.

Since $$E_{cell} = x \times 10^{-2}$$ V, we get $$x = \frac{1.4715}{10^{-2}} = 147.15$$.

Rounded to the nearest integer, $$x = 147$$.