For a cell involving one electron $E_{cell}^{\ominus} = 0.59 \; V$ at $298 \; K$,the equilibrium constant for the cell reaction is.
Given that $\frac{2.303 \; RT}{F} = 0.059 \; V$ at $T = 298 \; K$.

Which is the proper value of $x$ for the following to increase the cell potential of $Zn_{(s)} | Zn_{(x \ M)}^{2+} || Cu_{(0.02 \ M)}^{2+} | Cu_{(s)}$?

Write the Nernst equation and calculate the $emf$ of the following cells at $298 \, K$:
$(i) \; Mg_{(s)} | Mg^{2+}(0.001 \, M) || Cu^{2+}(0.0001 \, M) | Cu_{(s)}$
$(ii) \; Fe_{(s)} | Fe^{2+}(0.001 \, M) || H^{+}(1 \, M) | H_{2(g)}(1 \, bar) | Pt_{(s)}$
$(iii) \; Sn_{(s)} | Sn^{2+}(0.050 \, M) || H^{+}(0.020 \, M) | H_{2(g)}(1 \, bar) | Pt_{(s)}$
$(iv) \; Pt_{(s)} | Br_{2(l)} | Br^{-}(0.010 \, M), H^{+}(0.030 \, M) || H_{2(g)}(1 \, bar) | Pt_{(s)}$

For the cell at $298 \ K$:
$Ag_{(s)} | AgBr_{(s)} | Br^{-}(0.01 \ M) || I^{-}(0.02 \ M) | AgI_{(s)} | Ag_{(s)}$
The correct information is:
[Given: $K_{sp}(AgBr) = 4 \times 10^{-13}$,$K_{sp}(AgI) = 8 \times 10^{-17}$,$\frac{2.303 \ RT}{F} = 0.06 \ V$,$\log 2 = 0.3$]

For the cell reaction $Zn + Cu^{2+} \rightarrow Cu + Zn^{2+}$,the standard $EMF$ value at $25^{\circ}C$ is $1.10 \ V$. If $0.1 \ M \ Cu^{2+}$ and $0.1 \ M \ Zn^{2+}$ solutions are used,the $EMF$ will be .......... $V$.

The pressure of $H_2$ required to make the potential of $H_2$-electrode zero in pure water at $298 \ K$ is