The oxidation potentials of Zn, Cu, Ag, H_2 a | vedclass

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Question

(B) The cell voltage $(E_{cell})$ is calculated as $E_{cell} = E_{ox} + E_{red}$.For a reaction $A + B^{n+} \rightarrow A^{n+} + B$,$E_{cell} = E_{ox}

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$Zn + 2Ag^{+} \rightarrow 2Ag + Zn^{2+}$

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(B) The cell voltage $(E_{cell})$ is calculated as $E_{cell} = E_{ox} + E_{red}$.For a reaction $A + B^{n+} \rightarrow A^{n+} + B$,$E_{cell} = E_{ox}(A) + E_{red}(B^{n+})$.Given oxidation potentials $(E_{ox})$: $Zn = 0.76 \ V, Cu = -0.34 \ V, Ag = -0.80 \ V, H_2 = 0 \ V, Ni = 0.25 \ V$.Reduction potentials $(E_{red} = -E_{ox})$: $Cu^{2+} = 0.34 \ V, Ag^{+} = 0.80 \ V, H^{+} = 0 \ V, Ni^{2+} = -0.25 \ V$.$(A)$ $Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$: $E_{cell} = 0.76 + 0.34 = 1.10 \ V$.$(B)$ $Zn + 2Ag^{+} \rightarrow Zn^{2+} + 2Ag$: $E_{cell} = 0.76 + 0.80 = 1.56 \ V$.$(C)$ $H_2 + Cu^{2+} \rightarrow 2H^{+} + Cu$: $E_{cell} = 0 + 0.34 = 0.34 \ V$.$(D)$ $H_2 + Ni^{2+} \rightarrow 2H^{+} + Ni$: $E_{cell} = 0 + (-0.25) = -0.25 \ V$.The maximum voltage is $1.56 \ V$ for option $B$.

The oxidation potentials of $Zn, Cu, Ag, H_2$ and $Ni$ are $0.76, -0.34, -0.80, 0$ and $0.25 \ V$ respectively. Which reaction provides the maximum voltage?

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