Emf of the following cell at 298 ,K in V is x | vedclass

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(A) The cell reaction is: $Zn_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + 2Ag_{(s)}$$E^{0}_{cell} = E^{0}_{cathode} - E^{0}_{anode} = 0.80 - (

Vedclass · Accepted Answer

$147$

Vedclass · Answer

(A) The cell reaction is: $Zn_{(s)} + 2Ag^{+}_{(aq)} ightarrow Zn^{2+}_{(aq)} + 2Ag_{(s)}$$E^{0}_{cell} = E^{0}_{cathode} - E^{0}_{anode} = 0.80 - (-0.76) = 1.56 \,V$Using the Nernst equation: $E_{cell} = E^{0}_{cell} - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Ag^{+}]^{2}}$Here,$n = 2$,$[Zn^{2+}] = 0.1$,and $[Ag^{+}] = 0.01$$E_{cell} = 1.56 - \frac{0.059}{2} \log \frac{0.1}{(0.01)^{2}}$$E_{cell} = 1.56 - 0.0295 	imes \log(1000)$$E_{cell} = 1.56 - 0.0295 	imes 3 = 1.56 - 0.0885 = 1.4715 \,V$$E_{cell} = 147.15 	imes 10^{-2} \,V$Rounding to the nearest integer,$x = 147$.

$Emf$ of the following cell at $298 \,K$ in $V$ is $x \times 10^{-2}$. $Zn | Zn^{2+}(0.1 \,M) || Ag^{+}(0.01 \,M) | Ag$. The value of $x$ is .... . (Rounded off to the nearest integer) [Given: $E^{0}_{Zn^{2+}/Zn} = -0.76 \,V$; $E^{0}_{Ag^{+}/Ag} = +0.80 \,V$; $\frac{2.303 RT}{F} = 0.059$]

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