Chemistry · 2021
JEE Main · 26 February 2021, Shift 2 · Q57
Emf of the following cell at 298 K in V is x × 10^-2. Zn | Zn^2+(0.1 M ) ∥ Ag^+(0.01 M )| Ag The value of x is \ \ \ \. (Rounded off to the nearest…
Emf of the following cell at $\displaystyle 298$ K in V is $\displaystyle x \times 10^{-2}$.
$$\mathrm{Zn}\left|\mathrm{Zn}^{2+}(0.1 \mathrm{M}) \| \mathrm{Ag}^{+}(0.01 \mathrm{M})\right| \mathrm{Ag}
$$The value of $\displaystyle x$ is \_\_\_\_. (Rounded off to the nearest integer)
[Given : $\displaystyle \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^\theta=-0.76 \mathrm{~V} ; \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^\theta=+0.80 \mathrm{~V} ; \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059$ ]
Official answer
147
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