The standard reduction potentials for Zn^{2+} | vedclass

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Question

(A) reaction is spontaneous if the cell potential $E^0_{cell}$ is positive.For the reaction $X + Y^{2+} \rightarrow X^{2+} + Y$,the cell potential is

Vedclass · Accepted Answer

$X = Zn, Y = Ni$

Vedclass · Answer

(A) reaction is spontaneous if the cell potential $E^0_{cell}$ is positive.For the reaction $X + Y^{2+} \rightarrow X^{2+} + Y$,the cell potential is given by $E^0_{cell} = E^0_{cathode} - E^0_{anode} = E^0_{Y^{2+}/Y} - E^0_{X^{2+}/X}$.For the reaction to be spontaneous,$E^0_{Y^{2+}/Y} > E^0_{X^{2+}/X}$.Given values: $E^0_{Zn^{2+}/Zn} = -0.76 \, V$,$E^0_{Fe^{2+}/Fe} = -0.44 \, V$,$E^0_{Ni^{2+}/Ni} = -0.23 \, V$.Checking option $A$: $X = Zn, Y = Ni$. $E^0_{cell} = E^0_{Ni^{2+}/Ni} - E^0_{Zn^{2+}/Zn} = -0.23 - (-0.76) = +0.53 \, V$. Since $E^0_{cell} > 0$,the reaction is spontaneous.

The standard reduction potentials for $Zn^{2+}/Zn$,$Ni^{2+}/Ni$,and $Fe^{2+}/Fe$ are $-0.76 \, V$,$-0.23 \, V$,and $-0.44 \, V$ respectively. For which of the following combinations will the reaction $X + Y^{2+} \rightarrow X^{2+} + Y$ be spontaneous?

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