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(B) The cell reaction is: $2Cr_{(s)} + 3Fe^{2+}_{(aq)} \rightarrow 2Cr^{3+}_{(aq)} + 3Fe_{(s)}$The standard cell potential is: $E^0_{cell} = E^0_{cath

Vedclass · Accepted Answer

$0.26$

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(B) The cell reaction is: $2Cr_{(s)} + 3Fe^{2+}_{(aq)} ightarrow 2Cr^{3+}_{(aq)} + 3Fe_{(s)}$The standard cell potential is: $E^0_{cell} = E^0_{cathode} - E^0_{anode} = -0.42 - (-0.72) = 0.30 \, V$Using the Nernst equation at $298 \, K$: $E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log \frac{[Cr^{3+}]^2}{[Fe^{2+}]^3}$Here,$n = 6$ (total electrons transferred).$E_{cell} = 0.30 - \frac{0.0591}{6} \log \frac{(0.1)^2}{(0.01)^3}$$E_{cell} = 0.30 - 0.00985 \log \frac{10^{-2}}{10^{-6}} = 0.30 - 0.00985 \log(10^4)$$E_{cell} = 0.30 - (0.00985 	imes 4) = 0.30 - 0.0394 = 0.2606 \, V \approx 0.26 \, V$

Calculate the cell potential for the following cell: ............... $V$
$Cr_{(s)} | Cr^{3+}_{(0.1 \, M)} || Fe^{2+}_{(0.01 \, M)} | Fe_{(s)}$
Given: $E^0_{Cr^{3+}|Cr} = -0.72 \, V$,$E^0_{Fe^{2+}|Fe} = -0.42 \, V$ (in $, V$)

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Calculate the cell potential for the following cell: ............... $V$$Cr_{(s)} | Cr^{3+}_{(0.1 \, M)} || Fe^{2+}_{(0.01 \, M)} | Fe_{(s)}$Given: $E^0_{Cr^{3+}|Cr} = -0.72 \, V$,$E^0_{Fe^{2+}|Fe} = -0.42 \, V$ (in $, V$)