The electrode potential of the following half | vedclass

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(D) The cell reaction is: $X(s) + Y^{2+}(aq) \rightarrow X^{2+}(aq) + Y(s)$.The standard cell potential is $E^0_{cell} = E^0_{cathode} - E^0_{anode} =

Vedclass · Accepted Answer

$275$

Vedclass · Answer

(D) The cell reaction is: $X(s) + Y^{2+}(aq) ightarrow X^{2+}(aq) + Y(s)$.The standard cell potential is $E^0_{cell} = E^0_{cathode} - E^0_{anode} = 0.36 \ V - (-2.36 \ V) = 2.72 \ V$.Using the Nernst equation: $E_{cell} = E^0_{cell} - \frac{0.06}{n} \log \frac{[X^{2+}]}{[Y^{2+}]}$.Here,$n = 2$,$[X^{2+}] = 0.001 \ M$,and $[Y^{2+}] = 0.01 \ M$.$E_{cell} = 2.72 - \frac{0.06}{2} \log \frac{0.001}{0.01} = 2.72 - 0.03 \log(0.1) = 2.72 - 0.03(-1) = 2.72 + 0.03 = 2.75 \ V$.Thus,$E_{cell} = 275 	imes 10^{-2} \ V$.

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