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

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$1.927$

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(D) The cell reaction is: $3Fe(s) + 2Au^{+3}(aq) ightarrow 3Fe^{+2}(aq) + 2Au(s)$.The number of electrons transferred is $n = 6$.The standard cell potential is $E^o_{cell} = E^o_{cathode} - E^o_{anode} = 1.50\,V - (-0.44\,V) = 1.94\,V$.Using the Nernst equation: $E_{cell} = E^o_{cell} - \frac{0.0591}{n} \log \frac{[Fe^{+2}]^3}{[Au^{+3}]^2}$.$E_{cell} = 1.94 - \frac{0.0591}{6} \log \frac{(0.2)^3}{(0.02)^2}$.$E_{cell} = 1.94 - 0.00985 \log \frac{0.008}{0.0004} = 1.94 - 0.00985 \log(20)$.$E_{cell} = 1.94 - 0.00985 	imes 1.301 = 1.94 - 0.0128 = 1.9272\,V$.

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

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