Question

Calculate the emf of the cell $Cu/C{u}^{2+}\left(1M\right)//A{g}^{+}\left(0.01M\right)/Ag$ at ${25}^o$C. Given: $E^{o}Ag=0.8$V, $E^{o}Cu=0.34$V.

Answer 1

$Cu/C{u}^{+2}\left(1M\right)||A{g}^{+}\left(0.01M\right)/Ag$

$E^{o}A{g}^{+}/Ag=0.80V$

$E^{o}C{u}^{2+}/Cu=0.34V$

$E^{o}Cu/C{u}^{2+}=-0.34V$

$E_{cell}^o=E^{o}Cu/C{u}^{2+}+E^{o}A{g}^{+}/Ag$

$=-0.34+0.80=0.46V$

$Cu\to C{u}^{+2}+2e^{-}$

$[A{g}^{+}+e^{-}\to Ag]\times 2$

$Cu\to C{u}^{2+}+2e^{-}$

$2A{g}^{+}+2e^{-}\to Ag$

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$Cu+2A{g}^{+}\to C{u}^{2+}+Ag$

$E_{cell}=E_{cell}^o-\frac{0.0591}{n}log\frac{\left[P\right]}{\left[R\right]}$

$=0.46-\frac{0.0591}{2}log\frac{\left[C{u}^{+}2\right]}{[A{g}^{+}{]}^2}$

$=0.46-\frac{0.0591}{2}log10.01\times 0.01$

$=0.46-\frac{0.0591}{2}log{10}^4$

$=0.46-\frac{0.0591}{2}\left(4log10\right)$

$=0.46-\frac{0.0591}{2}\times 4$

$=0.46-\left(0.0591\right)\times 2$

$=0.3418V$