Consider the following cell reaction:2Fe_{(s) | vedclass

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(D) The given cell reaction is $2Fe_{(s)} + O_{2(g)} + 4H^{+}_{(aq)} 	o 2Fe^{2+}_{(aq)} + 2H_2O_{(l)}$.Here,$n = 4$ electrons are transferred.Given:

Vedclass · Accepted Answer

$1.57$

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(D) The given cell reaction is $2Fe_{(s)} + O_{2(g)} + 4H^{+}_{(aq)} 	o 2Fe^{2+}_{(aq)} + 2H_2O_{(l)}$.Here,$n = 4$ electrons are transferred.Given: $[Fe^{2+}] = 10^{-3} \ M$,$p(O_2) = 0.1 \ atm$,$pH = 3$,so $[H^{+}] = 10^{-3} \ M$.Using the Nernst equation:$E_{cell} = E^o_{cell} - \frac{0.0591}{n} \log \frac{[Fe^{2+}]^2}{[H^{+}]^4 \cdot p(O_2)}$$E_{cell} = 1.67 - \frac{0.0591}{4} \log \frac{(10^{-3})^2}{(10^{-3})^4 \cdot 0.1}$$E_{cell} = 1.67 - 0.014775 \cdot \log \frac{10^{-6}}{10^{-12} \cdot 10^{-1}}$$E_{cell} = 1.67 - 0.014775 \cdot \log 10^7$$E_{cell} = 1.67 - 0.014775 \cdot 7 = 1.67 - 0.1034 = 1.5666 \ V \approx 1.57 \ V$.

Consider the following cell reaction:
$2Fe_{(s)} + O_{2(g)} + 4H^{+}_{(aq)} \to 2Fe^{2+}_{(aq)} + 2H_2O_{(l)}$; $E^o = 1.67 \ V$
At $[Fe^{2+}] = 10^{-3} \ M$,$p(O_2) = 0.1 \ atm$ and $pH = 3$,the cell potential at $25 \ ^oC$ is .............. $V$.

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Consider the following cell reaction:$2Fe_{(s)} + O_{2(g)} + 4H^{+}_{(aq)} \to 2Fe^{2+}_{(aq)} + 2H_2O_{(l)}$; $E^o = 1.67 \ V$At $[Fe^{2+}] = 10^{-3} \ M$,$p(O_2) = 0.1 \ atm$ and $pH = 3$,the cell potential at $25 \ ^oC$ is .............. $V$.