Consider the following electrodes P = Zn^{2+} | vedclass

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(C) The standard reduction potential of $Zn^{2+} / Zn$ half-cell is $E^{\circ} = -0.76 \ V$. The Nernst equation for the reduction half-cell reaction

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$Q > R > S > P$

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(C) The standard reduction potential of $Zn^{2+} / Zn$ half-cell is $E^{\circ} = -0.76 \ V$. The Nernst equation for the reduction half-cell reaction $Zn^{2+}_{(aq)} + 2e^{-} \longrightarrow Zn_{(s)}$ is: $E_{red} = E^{\circ}_{red} - \frac{0.059}{n} \log \frac{1}{[Zn^{2+}]}$ $E_{red} = -0.76 + \frac{0.059}{2} \log [Zn^{2+}]$ Since the term $\frac{0.059}{2} \log [Zn^{2+}]$ is positive,a higher concentration of $Zn^{2+}$ leads to a higher (less negative) reduction potential. Comparing the concentrations: $[Q] = 0.1 \ M$,$[R] = 0.01 \ M$,$[S] = 0.001 \ M$,$[P] = 0.0001 \ M$. Therefore,the order of electrode potentials is $Q > R > S > P$.

Consider the following electrodes $P = Zn^{2+}(0.0001 \ M) / Zn$,$Q = Zn^{2+}(0.1 \ M) / Zn$,$R = Zn^{2+}(0.01 \ M) / Zn$,$S = Zn^{2+}(0.001 \ M) / Zn$. Given $E^{\circ}(Zn^{2+} / Zn) = -0.76 \ V$,the electrode potentials of the above electrodes in volts are in the order:

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