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(A) The cell reaction is $Zn_{(s)} + 2Ag^{+} \longrightarrow Zn^{2+} + 2Ag_{(s)}$.Here,$n = 2$ (number of electrons transferred).Using the Nernst equa

Vedclass · Accepted Answer

$2$

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(A) The cell reaction is $Zn_{(s)} + 2Ag^{+} \longrightarrow Zn^{2+} + 2Ag_{(s)}$.Here,$n = 2$ (number of electrons transferred).Using the Nernst equation at $298\, K$:$E_{	ext{cell}} = E_{	ext{cell}}^{\circ} - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Ag^{+}]^2}$Substituting the given values:$E_{	ext{cell}} = 2.57 - \frac{0.0591}{2} \log \frac{0.28}{(0.04)^2}$$E_{	ext{cell}} = 2.57 - 0.02955 \log \frac{0.28}{0.0016}$$E_{	ext{cell}} = 2.57 - 0.02955 \log(175)$Since $\log(175) \approx 2.243$,$E_{	ext{cell}} = 2.57 - 0.02955 	imes 2.243$$E_{	ext{cell}} = 2.57 - 0.0663 \approx 2.50\, V$.Thus,the correct option is $A$.

Consider the following electrochemical cell,$Zn_{(s)} + 2Ag^{+}(0.04\, M) \longrightarrow Zn^{2+}(0.28\, M) + 2Ag_{(s)}$. If $E_{\text{cell}}^{\circ} = 2.57\, V$,then the emf of the cell at $298\, K$ is $......\, V$. (in $.5$)

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