The maximum work that can be obtained from th | vedclass

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Question

(B) The cell reaction is:Anode: $X \rightarrow X^{2+} + 2e^-$Cathode: $2Y^+ + 2e^- \rightarrow 2Y$Overall reaction: $X + 2Y^+ \rightarrow X^{2+} + 2Y$

Vedclass · Accepted Answer

$482.5 \ kJ/mol$

Vedclass · Answer

(B) The cell reaction is:Anode: $X ightarrow X^{2+} + 2e^-$Cathode: $2Y^+ + 2e^- ightarrow 2Y$Overall reaction: $X + 2Y^+ ightarrow X^{2+} + 2Y$Here,the number of electrons involved,$n = 2$.$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.8 \ V - (-1.7 \ V) = 2.5 \ V$.The maximum work done is given by $W_{max} = -\Delta G^{\circ} = nFE^{\circ}_{cell}$.$W_{max} = 2 	imes 96500 \ C/mol 	imes 2.5 \ V = 482500 \ J/mol$.Converting to $kJ/mol$: $W_{max} = \frac{482500}{1000} = 482.5 \ kJ/mol$.

The maximum work that can be obtained from the following cell is
$X|X^{2+}_{(aq)} || Y^{+}_{(aq)}| Y$
Given,$E^{\circ}_{X^{2+}/X} = -1.7 \ V, E^{\circ}_{Y^{+}/Y} = 0.8 \ V$

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The maximum work that can be obtained from the following cell is$X|X^{2+}_{(aq)} || Y^{+}_{(aq)}| Y$Given,$E^{\circ}_{X^{2+}/X} = -1.7 \ V, E^{\circ}_{Y^{+}/Y} = 0.8 \ V$