Calculate the equilibrium constant of the following reaction:
$\text{Cu(s)} + 2\text{Ag}^+_{\text{(aq)}} \rightarrow \text{Cu}^{2+}_{\text{(aq)}} + 2\text{Ag(s)}$, given $E^\circ_{\text{cell}} = 0.46 \text{ V}$.

Consider the cell $Pt | H_2(P_1 \ atm) | H^{+}(X_1 \ M) || H^{+}(X_2 \ M) | H_2(P_2 \ atm) | Pt$. The cell reaction will be spontaneous if

Calculate the cell potential for $Zn_{(s)} | Zn^{2+} (0.6 \ M) || Cd^{2+} (0.2 \ M) | Cd_{(s)}$ at $298 \ K$. (in $V$)

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

Which one of the following will increase the voltage of the cell? $(T = 298 \ K)$ :- $Sn_{(s)} + 2Ag_{(aq)}^{+} \rightarrow Sn_{(aq)}^{2+} + 2Ag_{(s)}$

The $emf$ of a $Daniel$ cell at $298 \ K$ is ${E_1}$ for the cell reaction $Zn|ZnSO_4(0.01 \ M)||CuSO_4(1.0 \ M)|Cu$. When the concentration of $ZnSO_4$ is $1.0 \ M$ and that of $CuSO_4$ is $0.01 \ M$,the $emf$ changes to ${E_2}$. What is the relationship between ${E_1}$ and ${E_2}$?