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

Vedclass · Accepted Answer

$0.32 / 10^{0.0295}$

Vedclass · Answer

(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.

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 a galvanic cell reaction at $25^{\circ}C$ with $n = 4$,the standard $emf$ is $0.295 \ V$. What is the equilibrium constant for the reaction? $(F = 96500 \ C \ mol^{-1}; R = 8.314 \ J \ K^{-1} \ mol^{-1})$

If $Zn/Zn^{+2}$ electrode is diluted $100$ times,then the change in electromotive force will be:

The cell potential for the following cell notation is approximately
$M_{(s)} | M^{3+}(aq, 0.01 \ M) || N^{2+}(aq, 0.1 \ M) | N_{(s)}$
$E_{M^{3+} / M}^0 = 0.6 \ V$ and $E_{N^{2+} / N}^0 = 0.1 \ V$ (in $V$)

Identify the correct statement$(s)$:

For the cell reaction $Zn + Cu^{2+} \rightarrow Cu + Zn^{2+}$,the standard $EMF$ value at $25^{\circ}C$ is $1.10 \ V$. If $0.1 \ M \ Cu^{2+}$ and $0.1 \ M \ Zn^{2+}$ solutions are used,the $EMF$ will be .......... $V$.

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