For the cell $Cu_{(s)}|Cu^{2+}_{(aq)}(0.1 \ M) || Ag^{+}_{(aq)}(0.01 \ M)| Ag_{(s)}$,the cell potential $E_{1} = 0.3095 \ V$. For the cell $Cu_{(s)}|Cu^{2+}_{(aq)}(0.01 \ M) || Ag^{+}_{(aq)}(0.001 \ M)| Ag_{(s)}$,the cell potential $= ..... \times 10^{-2} \ V$. (Round off to the Nearest Integer). [Use: $\frac{2.303 \ RT}{F} = 0.059$]

Calculate the cell potential at $298 \ K$ for the following cell:
$Zn_{(s)} | Zn^{2+} (0.6 \ M) || Cu^{2+} (0.3 \ M) | Cu_{(s)} \quad [E_{cell}^{o} = 1.1 \ V]$

What is the oxidation potential of $0.05 \, M \, H_2SO_4$ in volts?

For the electrochemical cell,$Mg_{(s)} \mid Mg^{2+}(aq, 1 \ M) \parallel Cu^{2+}(aq, 1 \ M) \mid Cu_{(s)}$,the standard emf of the cell is $2.70 \ V$ at $300 \ K$. When the concentration of $Mg^{2+}$ is changed to $x$,the cell potential changes to $2.67 \ V$ at $300 \ K$. The value of $x$ is.
(Given: $\frac{F}{R} = 11500 \ K \ V^{-1}$,where $F$ is the Faraday constant and $R$ is the gas constant; $\ln(10) = 2.30$)

For a cell involving a one-electron change at $25^o C$,$E^{o}_{cell} = 0.591 \ V$. The equilibrium constant for the reaction is .....

For the cell $Fe | Fe^{+2} (x \, M) || Cu^{+2} (0.01 \, M) | Cu$,if $E_{cell} = 0.78 \, V$,$E^{\circ}_{Fe^{+2} | Fe} = -0.44 \, V$,and $E^{\circ}_{Cu^{+2} | Cu} = +0.34 \, V$,determine the value of $x$.