Identify which of the following cell reaction | vedclass

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Question

(A) reaction is spontaneous if the standard cell potential $E^0_{cell} > 0$. The formula is $E^0_{cell} = E^0_{cathode} - E^0_{anode}$.For $(A)$: $Ca$

Vedclass · Accepted Answer

$Ca(s) + Cd^{2+}(aq) ightarrow Ca^{2+}(aq) + Cd(s), [E^0_{Ca^{2+}/Ca} = -2.866 	ext{ V}, E^0_{Cd^{2+}/Cd} = -0.403 	ext{ V}]$

Vedclass · Answer

(A) reaction is spontaneous if the standard cell potential $E^0_{cell} > 0$. The formula is $E^0_{cell} = E^0_{cathode} - E^0_{anode}$.For $(A)$: $Ca$ is oxidized (anode), $Cd^{2+}$ is reduced (cathode). $E^0_{cell} = (-0.403) - (-2.866) = +2.463 	ext{ V}$. Since $E^0_{cell} > 0$, the reaction is spontaneous.For $(B)$: $Br^-$ is oxidized (anode), $Sn^{2+}$ is reduced (cathode). $E^0_{cell} = (-0.136) - (1.08) = -1.216 	ext{ V}$. Non-spontaneous.For $(C)$: $Ag$ is oxidized (anode), $Ni^{2+}$ is reduced (cathode). $E^0_{cell} = (-0.257) - (0.799) = -1.056 	ext{ V}$. Non-spontaneous.For $(D)$: $Au$ is oxidized (anode), $Zn^{2+}$ is reduced (cathode). $E^0_{cell} = (-0.763) - (1.68) = -2.443 	ext{ V}$. Non-spontaneous.Therefore, option $(A)$ is correct.

Identify which of the following cell reactions is spontaneous under standard state conditions.

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