For the cell reaction,2Ce^{4+} + Co to 2Ce^{3 | vedclass

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Question

(B) The cell reaction is $2Ce^{4+} + Co 	o 2Ce^{3+} + Co^{2+}$.In this reaction,$Co$ is oxidized to $Co^{2+}$ (anode) and $Ce^{4+}$ is reduced to $Ce

Vedclass · Accepted Answer

$+1.64 \ V$

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(B) The cell reaction is $2Ce^{4+} + Co 	o 2Ce^{3+} + Co^{2+}$.In this reaction,$Co$ is oxidized to $Co^{2+}$ (anode) and $Ce^{4+}$ is reduced to $Ce^{3+}$ (cathode).$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$$E^\circ_{cell} = E^\circ_{Ce^{4+}/Ce^{3+}} - E^\circ_{Co^{2+}/Co}$Given $E^\circ_{cell} = 1.89 \ V$ and $E^\circ_{Co^{2+}/Co} = -0.28 \ V$.$1.89 = E^\circ_{Ce^{4+}/Ce^{3+}} - (-0.28)$$E^\circ_{Ce^{4+}/Ce^{3+}} = 1.89 - 0.28 = 1.61 \ V$.The closest option provided is $1.64 \ V$.

For the cell reaction,$2Ce^{4+} + Co \to 2Ce^{3+} + Co^{2+}$,$E^\circ_{cell}$ is $1.89 \ V$. If $E^\circ_{Co^{2+}/Co} = -0.28 \ V$,then $E^\circ_{Ce^{4+}/Ce^{3+}}$ is equal to:

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