Calculate the emf of the cell: Cr | Cr^{3+}(0 | vedclass

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(A) The standard cell potential $E^0_{cell}$ is calculated as $E^0_{cathode} - E^0_{anode}$.$E^0_{cell} = E^0_{Fe^{2+}/Fe} - E^0_{Cr^{3+}/Cr} = -0.45

Vedclass · Accepted Answer

$0.2606 \ V$

Vedclass · Answer

(A) The standard cell potential $E^0_{cell}$ is calculated as $E^0_{cathode} - E^0_{anode}$.$E^0_{cell} = E^0_{Fe^{2+}/Fe} - E^0_{Cr^{3+}/Cr} = -0.45 \ V - (-0.75 \ V) = 0.30 \ V$.The Nernst equation at $298 \ K$ is $E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log \frac{[Cr^{3+}]^2}{[Fe^{2+}]^3}$.Here,$n = 6$ (number of electrons transferred).$E_{cell} = 0.30 - \frac{0.0591}{6} \log \frac{(0.1)^2}{(0.01)^3} = 0.30 - 0.00985 \log \frac{10^{-2}}{10^{-6}} = 0.30 - 0.00985 \log(10^4)$.$E_{cell} = 0.30 - 0.00985 	imes 4 = 0.30 - 0.0394 = 0.2606 \ V$.

Calculate the $emf$ of the cell: $Cr | Cr^{3+}(0.1 \ M) || Fe^{2+}(0.01 \ M) | Fe$. Given: $E^0_{Cr^{3+}/Cr} = -0.75 \ V$; $E^0_{Fe^{2+}/Fe} = -0.45 \ V$. Cell reaction: $2 \ Cr_{(s)} + 3 \ Fe^{2+}_{(aq)} \rightarrow 2 \ Cr^{3+}_{(aq)} + 3 \ Fe_{(s)}$.

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