If the standard reduction potentials of Zn, N | vedclass

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(D) For a reaction to be spontaneous, the standard cell potential $E^\circ_{	ext{cell}}$ must be positive.$E^\circ_{	ext{cell}} = E^\circ_{	ext{cat

Vedclass · Accepted Answer

$X = Zn, Y = Ni$

Vedclass · Answer

(D) For a reaction to be spontaneous, the standard cell potential $E^\circ_{	ext{cell}}$ must be positive.$E^\circ_{	ext{cell}} = E^\circ_{	ext{cathode}} - E^\circ_{	ext{anode}} = E^\circ_{Y^{+2}/Y} - E^\circ_{X^{+2}/X} > 0$.This implies $E^\circ_{Y^{+2}/Y} > E^\circ_{X^{+2}/X}$.Given potentials: $E^\circ_{Zn^{+2}/Zn} = -0.76 	ext{ V}$, $E^\circ_{Fe^{+2}/Fe} = -0.44 	ext{ V}$, $E^\circ_{Ni^{+2}/Ni} = -0.23 	ext{ V}$.Checking option $(D)$: $X = Zn$ $(E^\circ = -0.76 	ext{ V})$ and $Y = Ni$ $(E^\circ = -0.23 	ext{ V})$.$E^\circ_{	ext{cell}} = (-0.23) - (-0.76) = +0.53 	ext{ V}$.Since $E^\circ_{	ext{cell}} > 0$, the reaction is spontaneous.

If the standard reduction potentials of $Zn$, $Ni$, and $Fe$ are $-0.76 \text{ V}$, $-0.23 \text{ V}$, and $-0.44 \text{ V}$ respectively, determine the electrodes $X$ and $Y$ for the reaction $X(s) + Y^{+2}_{(aq)} \rightarrow X^{+2}_{(aq)} + Y(s)$ to be spontaneous.

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