For the cell $Zn_{(s)} | Zn^{2+} (0.1 \ M) || Fe^{2+} (0.01 \ M) | Fe_{(s)}$ at $298 \ K$,$E_{cell} = 0.2905 \ V$. What is the equilibrium constant $(K_c)$ for the reaction $Zn_{(s)} + Fe^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Fe_{(s)}$?

The Gibbs energy change of the reaction (in $kJ \ mol^{-1}$) corresponding to the following cell $Cr | Cr^{3+} (0.1 \ M) || Fe^{2+} (0.01 \ M) | Fe$ is: (Given: $E^{\circ}_{Cr^{3+}/Cr} = -0.74 \ V$,$E^{\circ}_{Fe^{2+}/Fe} = -0.44 \ V$)

The $E^o$ for the cell $Zn|Zn^{2+}_{(aq)}||Cu^{2+}_{(aq)}|Cu$ is $1.10 \ V$ at $25 \ ^oC$. The equilibrium constant for the reaction $Zn + Cu^{2+}_{(aq)} \rightleftharpoons Cu + Zn^{2+}_{(aq)}$ is of the order of:

For the cell reaction $Zn_{(s)} + 2H^+_{(aq)} \to Zn^{2+}_{(aq)} + H_{2(g)}$,what happens when $H_2SO_4$ is added to the cathode compartment?

One half cell in a voltaic cell is constructed by dipping a silver rod in an $AgNO_3$ solution of unknown concentration, and the other half cell is a $Zn$ rod dipped in a $1 \text{ M}$ solution of $ZnSO_4$. $A$ voltage of $1.60 \text{ V}$ is measured at $298 \text{ K}$ for this cell. What is the concentration of $Ag^+$ ions in terms of $\log x$ (where $x = [Ag^+]$)? Given: $E^\ominus_{Zn^{2+}/Zn} = -0.76 \text{ V}$, $E^\ominus_{Ag^+/Ag} = +0.80 \text{ V}$, and $\frac{2.303RT}{F} = 0.059 \text{ V}$.

If the $E^{\circ}_{cell}$ of an equilibrium reaction $A_{(s)} + 2B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + 2B_{(s)}$ at $298 \ K$ is $0.59 \ V$,the equilibrium constant $K_c$ is