Calculate the cell potential for the cell Zn_ | vedclass

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Question

(A) The cell reaction is: $Zn_{(s)} + Cd^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cd_{(s)}$Standard cell potential $E^{\circ}_{cell} = E^{\circ}_{cath

Vedclass · Accepted Answer

$0.36$

Vedclass · Answer

(A) The cell reaction is: $Zn_{(s)} + Cd^{2+}_{(aq)} ightarrow Zn^{2+}_{(aq)} + Cd_{(s)}$Standard cell potential $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = -0.40 \ V - (-0.76 \ V) = 0.36 \ V$.Using the Nernst equation at $298 \ K$: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Cd^{2+}]}$.Here $n = 2$,$[Zn^{2+}] = 0.6 \ M$,and $[Cd^{2+}] = 0.85 \ M$.$E_{cell} = 0.36 - \frac{0.0591}{2} \log \frac{0.6}{0.85}$.$E_{cell} = 0.36 - 0.02955 	imes \log(0.7059)$.$E_{cell} = 0.36 - 0.02955 	imes (-0.1513) = 0.36 + 0.00447 = 0.36447 \ V$.Rounding to two decimal places,the cell potential is $0.36 \ V$.

Calculate the cell potential for the cell $Zn_{(s)} | Zn^{2+} (0.6 \ M) || Cd^{2+} (0.85 \ M) | Cd_{(s)}$ at $298 \ K$. (Given: $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$ and $E^{\circ}_{Cd^{2+}/Cd} = -0.40 \ V$) (in $V$)

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