Calculate the equilibrium constant of the reaction:
$Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$
Given $E^{\Theta}_{cell} = 0.46 \ V$

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

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

Consider the following electrochemical cell at $298 \ K$:
$Pt | HSnO_2^-(aq) | Sn(OH)_6^{2-}(aq) || Bi_2O_3(s) | Bi(s)$.
If the reaction quotient at a given time is $10^6$, then the cell $EMF$ $(E_{\text{cell}})$ is . . . . . . $\times 10^{-1} \ V$ (Nearest integer).
Given the standard half-cell reduction potential as
$E^0_{Bi_2O_3/Bi, OH^-} = -0.44 \ V$ and
$E^0_{Sn(OH)_6^{2-}/HSnO_2^-, OH^-} = -0.90 \ V$.

The standard electrode potential for $Cu^{+2}/Cu$ is $0.34 \ V$. Calculate the reduction potential at $pH = 14$ for the above couple $V$ $[K_{sp}[Cu(OH)_2] = 1 \times 10^{-19}]$

For the cell reaction $Zn_{(s)} + Ni^{2+}_{(aq)} (a = 0.1) \rightarrow Zn^{2+}_{(aq)} (a = 1.0) + Ni_{(s)}$ at $298 \ K$,the $emf$ is observed to be $0.5105 \ V$. What is the standard $emf$ $(E^{\circ}_{cell})$ of the cell in $V$?