Calculate the E_{cell}^o for Zn_{(s)}|Zn_{(1M | vedclass

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Question

(A) The cell reaction is represented as $Zn_{(s)}|Zn^{2+}_{(1M)}||Cd^{2+}_{(1M)}|Cd_{(s)}$.In this cell,$Zn$ acts as the anode (oxidation) and $Cd$ ac

Vedclass · Accepted Answer

$0.36$

Vedclass · Answer

(A) The cell reaction is represented as $Zn_{(s)}|Zn^{2+}_{(1M)}||Cd^{2+}_{(1M)}|Cd_{(s)}$.In this cell,$Zn$ acts as the anode (oxidation) and $Cd$ acts as the cathode (reduction).The standard cell potential is calculated using the formula:$E_{cell}^o = E_{cathode}^{\circ} - E_{anode}^{\circ}$Substituting the given values:$E_{cell}^o = E_{Cd^{2+}/Cd}^{\circ} - E_{Zn^{2+}/Zn}^{\circ}$$E_{cell}^o = -0.403 \ V - (-0.763 \ V)$$E_{cell}^o = -0.403 \ V + 0.763 \ V = 0.36 \ V$.

Calculate the $E_{cell}^o$ for $Zn_{(s)}|Zn_{(1M)}^{2+}| |Cd_{(1M)}^{2+}|Cd_{(s)}$ at $25^{\circ} C$ given that $E_{Zn^{2+}/Zn}^{\circ} = -0.763 \ V$ and $E_{Cd^{2+}/Cd}^{\circ} = -0.403 \ V$. (in $V$)

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