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

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Question

(B) The cell reaction is $2Ce^{4+} + Co 	o 2Ce^{3+} + Co^{2+}$.Anode reaction: $Co 	o Co^{2+} + 2e^-$,so $E_{anode}^o = E_{Co^{2+}/Co}^o = -E_{Co/Co

Vedclass · Accepted Answer

$+1.61$

Vedclass · Answer

(B) The cell reaction is $2Ce^{4+} + Co 	o 2Ce^{3+} + Co^{2+}$.Anode reaction: $Co 	o Co^{2+} + 2e^-$,so $E_{anode}^o = E_{Co^{2+}/Co}^o = -E_{Co/Co^{2+}}^o = -0.28 \, V$.Cathode reaction: $2Ce^{4+} + 2e^- 	o 2Ce^{3+}$,so $E_{cathode}^o = E_{Ce^{4+}/Ce^{3+}}^o = x$.The standard cell potential is given by $E_{cell}^o = E_{cathode}^o - E_{anode}^o$.Substituting the values: $1.89 = x - (-0.28)$.$1.89 = x + 0.28$.$x = 1.89 - 0.28 = 1.61 \, V$.

For the cell reaction,$2Ce^{4+} + Co \to 2Ce^{3+} + Co^{2+}$,$E_{cell}^o$ is $1.89 \, V$ and $E_{Co/Co^{2+}}^o = +0.28 \, V$. If $E_{Ce^{4+}/Ce^{3+}}^o = x$,then $x$ is ............. $V$.

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