What will be the electromotive force of the f | vedclass

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(D) The cell reaction is: $3Fe(s) + 2Au^{+3}(aq) \rightarrow 3Fe^{+2}(aq) + 2Au(s)$Number of electrons transferred $(n)$ = $6$.Standard cell potential

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$1.927$

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(D) The cell reaction is: $3Fe(s) + 2Au^{+3}(aq) ightarrow 3Fe^{+2}(aq) + 2Au(s)$Number of electrons transferred $(n)$ = $6$.Standard cell potential: $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = 1.50 \ V - (-0.44 \ V) = 1.94 \ V$.Using the Nernst equation: $E_{cell} = E_{cell}^{\circ} - \frac{0.0591}{n} \log \frac{[Fe^{+2}]^3}{[Au^{+3}]^2}$.$E_{cell} = 1.94 - \frac{0.06}{6} \log \frac{(0.2)^3}{(0.02)^2}$.$E_{cell} = 1.94 - 0.01 \log \frac{0.008}{0.0004} = 1.94 - 0.01 \log(20)$.$E_{cell} = 1.94 - 0.01 	imes 1.301 \approx 1.94 - 0.013 = 1.927 \ V$.

What will be the electromotive force of the following cell in $V$?
$Fe|Fe^{+2}(0.2 \ M)||Au^{+3}(0.02 \ M)|Au$
Given: $E_{Fe^{+2}/Fe}^o = -0.44 \ V$ and $E_{Au^{+3}/Au}^o = 1.50 \ V$

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What will be the electromotive force of the following cell in $V$?$Fe|Fe^{+2}(0.2 \ M)||Au^{+3}(0.02 \ M)|Au$Given: $E_{Fe^{+2}/Fe}^o = -0.44 \ V$ and $E_{Au^{+3}/Au}^o = 1.50 \ V$