Find the observed EMF of the cell Cd | Cd^{2+ | vedclass

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(B) The cell reaction is $Cd_{(s)} + Cu^{2+}(aq) \longrightarrow Cd^{2+}(aq) + Cu_{(s)}$.First,calculate the standard cell potential: $E^{\circ}_{cell

Vedclass · Accepted Answer

$0.15$

Vedclass · Answer

(B) The cell reaction is $Cd_{(s)} + Cu^{2+}(aq) \longrightarrow Cd^{2+}(aq) + Cu_{(s)}$.First,calculate the standard cell potential: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.35 \ V - (-0.40 \ V) = 0.75 \ V$.Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log Q$.Here,$Q = \frac{[Cd^{2+}]}{[Cu^{2+}]} = \frac{0.01}{0.01} = 1$.Since $\log(1) = 0$,$E_{cell} = 0.75 \ V$.The observed $EMF$ under load is given by $E_{observed} = E_{cell} - (I 	imes R)$.$E_{observed} = 0.75 \ V - (0.15 \ A 	imes 4 \ \Omega) = 0.75 \ V - 0.60 \ V = 0.15 \ V$.

Find the observed $EMF$ of the cell $Cd | Cd^{2+}(0.01 \ M) || Cu^{2+}(0.01 \ M) | Cu$ under conditions where the internal resistance is $4 \ \Omega$ and it is producing a current of $0.15 \ A$. (Given: $E^{\circ}_{Cu^{2+}/Cu} = 0.35 \ V$ and $E^{\circ}_{Cd^{2+}/Cd} = -0.4 \ V$) (in $V$)

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