Calculate the standard electromotive force (E | vedclass

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(A) The cell reaction is represented as $Zn|Zn^{2+}||CuI|Cu$.At the anode (oxidation): $Zn_{(s)} 	o Zn^{2+}_{(aq)} + 2e^{-}$,$E^o_{ox} = -(-0.76 \ V)

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$0.59$

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(A) The cell reaction is represented as $Zn|Zn^{2+}||CuI|Cu$.At the anode (oxidation): $Zn_{(s)} 	o Zn^{2+}_{(aq)} + 2e^{-}$,$E^o_{ox} = -(-0.76 \ V) = +0.76 \ V$.At the cathode (reduction): $2CuI_{(s)} + 2e^{-} 	o 2Cu_{(s)} + 2I^{-}_{(aq)}$,$E^o_{red} = -0.17 \ V$.The overall cell reaction is $Zn_{(s)} + 2CuI_{(s)} 	o Zn^{2+}_{(aq)} + 2Cu_{(s)} + 2I^{-}_{(aq)}$.The standard cell potential is $E^o_{cell} = E^o_{cathode} - E^o_{anode} = E^o_{red(cathode)} - E^o_{red(anode)}$.$E^o_{cell} = -0.17 \ V - (-0.76 \ V) = -0.17 \ V + 0.76 \ V = 0.59 \ V$.

Calculate the standard electromotive force $(E^o_{cell})$ for the following cell: $Zn|Zn^{2+} (1 \ M)||I^{-} (1 \ M)|CuI|Cu$. Given the half-cell reactions:
$CuI_{(s)} + e^{-} \to Cu_{(s)} + I^{-}_{(aq)}$ ; $E^o = -0.17 \ V$
$Zn^{2+}_{(aq)} + 2e^{-} \to Zn_{(s)}$ ; $E^o = -0.76 \ V$ (in $V$)

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Calculate the standard electromotive force $(E^o_{cell})$ for the following cell: $Zn|Zn^{2+} (1 \ M)||I^{-} (1 \ M)|CuI|Cu$. Given the half-cell reactions:$CuI_{(s)} + e^{-} \to Cu_{(s)} + I^{-}_{(aq)}$ ; $E^o = -0.17 \ V$$Zn^{2+}_{(aq)} + 2e^{-} \to Zn_{(s)}$ ; $E^o = -0.76 \ V$ (in $V$)