Calculate the cell potential for Cr_{(s)} | C | vedclass

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Question

(A) The cell reaction is: $2Cr_{(s)} + 3Fe^{2+}_{(aq)} \rightarrow 2Cr^{3+}_{(aq)} + 3Fe_{(s)}$.$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{a

Vedclass · Accepted Answer

$0.26$

Vedclass · Answer

(A) The cell reaction is: $2Cr_{(s)} + 3Fe^{2+}_{(aq)} ightarrow 2Cr^{3+}_{(aq)} + 3Fe_{(s)}$.$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = -0.44 \, V - (-0.74 \, V) = 0.30 \, V$.Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Cr^{3+}]^2}{[Fe^{2+}]^3}$.Here $n = 6$,$[Cr^{3+}] = 0.1 \, M$,and $[Fe^{2+}] = 0.01 \, M$.$E_{cell} = 0.30 - \frac{0.0591}{6} \log \frac{(0.1)^2}{(0.01)^3} = 0.30 - 0.00985 \log \frac{10^{-2}}{10^{-6}} = 0.30 - 0.00985 \log(10^4)$.$E_{cell} = 0.30 - 0.00985 	imes 4 = 0.30 - 0.0394 = 0.2606 \, V \approx 0.26 \, V$.

Calculate the cell potential for $Cr_{(s)} | Cr^{3+} (0.1 \, M) || Fe^{2+} (0.01 \, M) | Fe_{(s)}$ at $298 \, K$. Given $E^{\circ}_{Cr^{3+}/Cr} = -0.74 \, V$ and $E^{\circ}_{Fe^{2+}/Fe} = -0.44 \, V$. (in $, V$)

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