At 298 , K,find out the emf for the cell: Al_ | vedclass

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(B) The cell reaction is: $2 Al_{(s)} + 3 Fe^{+2} (aq) \longrightarrow 2 Al^{+3} (aq) + 3 Fe_{(s)}$$E^o_{	ext{cell}} = E^o_{	ext{cathode}} - E^o_{

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$1.15$

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(B) The cell reaction is: $2 Al_{(s)} + 3 Fe^{+2} (aq) \longrightarrow 2 Al^{+3} (aq) + 3 Fe_{(s)}$$E^o_{	ext{cell}} = E^o_{	ext{cathode}} - E^o_{	ext{anode}} = -0.44 - (-1.66) = 1.22 \, V$Using the Nernst equation at $298 \, K$:$E_{	ext{cell}} = E^o_{	ext{cell}} - \frac{0.0591}{n} \log \frac{[Al^{+3}]^2}{[Fe^{+2}]^3}$Here,$n = 6$ (total electrons transferred).$E_{	ext{cell}} = 1.22 - \frac{0.06}{6} \log \frac{(0.1)^2}{(0.001)^3}$$E_{	ext{cell}} = 1.22 - 0.01 \log \frac{10^{-2}}{10^{-9}} = 1.22 - 0.01 \log(10^7)$$E_{	ext{cell}} = 1.22 - 0.01 	imes 7 = 1.22 - 0.07 = 1.15 \, V$

At $298 \, K$,find out the $emf$ for the cell: $Al_{(s)} | Al^{+3} (0.1 \, M) || Fe^{+2} (0.001 \, M) | Fe_{(s)}$. Given $E^o_{Al^{+3}/Al} = -1.66 \, V$ and $E^o_{Fe^{+2}/Fe} = -0.44 \, V$. (in $, V$)

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