The e.m.f. of the cell Zn | Zn^{2+} (0.01 M) | vedclass

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(B) The cell reaction is: $Zn(s) + Fe^{2+}(aq) 	o Zn^{2+}(aq) + Fe(s)$.Using the Nernst equation: $E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log \fra

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$0.32 / 10^{0.0295}$

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(B) The cell reaction is: $Zn(s) + Fe^{2+}(aq) 	o Zn^{2+}(aq) + Fe(s)$.Using the Nernst equation: $E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Fe^{2+}]}$.Given $E_{cell} = 0.2905 \ V$,$n = 2$,$[Zn^{2+}] = 0.01 \ M$,and $[Fe^{2+}] = 0.001 \ M$.$0.2905 = E^0_{cell} - \frac{0.0591}{2} \log \frac{10^{-2}}{10^{-3}} = E^0_{cell} - 0.02955 \log(10) = E^0_{cell} - 0.02955$.$E^0_{cell} = 0.2905 + 0.02955 = 0.32005 \approx 0.32 \ V$.At equilibrium,$E_{cell} = 0$,so $E^0_{cell} = \frac{0.0591}{n} \log K_c$.$0.32 = \frac{0.0591}{2} \log K_c = 0.02955 \log K_c$.$\log K_c = \frac{0.32}{0.02955} \approx \frac{0.32}{0.0295}$.Therefore,$K_c = 10^{\frac{0.32}{0.0295}}$. Note: The provided option $B$ is represented as $10^{0.32/0.0295}$ in standard notation,but based on the provided options,$B$ is the intended choice.

$A$. $Ag^{+}(0.0001 \ M) / Ag_{(s)}$	$B$. $Ag^{+}(0.1 \ M) / Ag_{(s)}$
$C$. $Ag^{+}(0.01 \ M) / Ag_{(s)}$	$D$. $Ag^{+}(0.001 \ M) / Ag_{(s)}$

The e.m.f. of the cell $Zn | Zn^{2+} (0.01 \ M) || Fe^{2+} (0.001 \ M) | Fe$ at $298 \ K$ is $0.2905 \ V$. The value of the equilibrium constant for the cell reaction is:

Similar Questions

For the cell $Zn_{(s)} | Zn^{2+} (0.1 \ M) || Fe^{2+} (0.01 \ M) | Fe_{(s)}$ at $298 \ K$,$E_{cell} = 0.2905 \ V$. What is the equilibrium constant $(K_c)$ for the reaction $Zn_{(s)} + Fe^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Fe_{(s)}$?

Consider the following $4$ electrodes:
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$C$. $Ag^{+}(0.01 \ M) / Ag_{(s)}$ $D$. $Ag^{+}(0.001 \ M) / Ag_{(s)}$

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