For the cell $Zn | Zn^{2+} (1 \ M) || Cu^{2+} (1 \ M) | Cu$ $(E^{\circ}_{cell} = 1.10 \ V)$ at $298 \ K$,when the cell is completely discharged,what is the ratio of concentrations $\frac{[Zn^{2+}]}{[Cu^{2+}]}$?

At $298 \ K$ the equilibrium constant for the reaction $M_{(s)} + 2 Ag^{+}_{(aq)} \rightarrow M^{2+}_{(aq)} + 2 Ag_{(s)}$ is $10^{15}$. What is the $E_{cell}^{\ominus}$ (in $V$) for this reaction? $\left(\frac{2.303 RT}{F}\right) = 0.06 \ V$

The reduction potential of a hydrogen half-cell will be negative if:

The reduction potential of a hydrogen half-cell will be positive if:

For the cell $Zn_{(s)} | Zn^{2+}_{(1\,M)} || Cu^{2+}_{(1\,M)} | Cu_{(s)}$,the $E^o_{cell} = 1.10 \, V$. When the cell is fully discharged at $298 \, K$,the ratio of the concentrations $[Zn^{2+}] / [Cu^{2+}]$ is:

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: