The cell potential for the following reaction | vedclass

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(A) The cell reaction is: $Zn(s) + Cd^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cd(s)$.$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = -0.40 \

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$0.01$

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(A) The cell reaction is: $Zn(s) + Cd^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cd(s)$.$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = -0.40 \ V - (-0.76 \ V) = 0.36 \ V$.Using the Nernst equation at $298 \ K$: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Cd^{2+}]}$.Here,$n = 2$,$E_{cell} = 0.03305 \ V$,$[Zn^{2+}] = 0.1 \ M$,$[Cd^{2+}] = x$.$0.03305 = 0.36 - \frac{0.0591}{2} \log \frac{0.1}{x}$.$0.36 - 0.03305 = 0.02955 \log \frac{0.1}{x}$.$0.32695 = 0.02955 \log \frac{0.1}{x}$.$\log \frac{0.1}{x} = \frac{0.32695}{0.02955} \approx 11.06$.This calculation suggests a discrepancy in the provided potential value or standard values. Assuming the standard textbook problem context where $E_{cell} = 0.36 - 0.02955 \log(0.1/x)$,for $x = 0.01 \ M$,$E_{cell} = 0.36 - 0.02955 \log(10) = 0.36 - 0.02955 = 0.33045 \ V \approx 0.3305 \ V$. Thus,$x = 0.01 \ M$.

The cell potential for the following reaction is $0.03305 \ V$ at $298 \ K$. Find the value of $x$ for the reaction: $Zn | Zn^{2+} (0.1 \ M) || Cd^{2+} (x \ M) | Cd$. (Given: $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$,$E^{\circ}_{Cd^{2+}/Cd} = -0.40 \ V$) (in $M$)

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