Calculate the $emf$ of the cell at $25^{\circ} C$.
Cell notation: $M | M^{2+} (0.01 \ M) || M^{2+} (0.0001 \ M) | M$
Given: $E_{cell}^{o} = 4 \ V$ and $\frac{RT}{F} \ln 10 = 0.06$. (in $V$)

For the cell $Zn | Zn^{2+}(0.01 \, M) || Fe^{2+}(0.001 \, M) | Fe$ at $25^o C$,the $E_{cell} = 0.2905 \, V$. The equilibrium constant $K_c$ is:

For the reduction of silver ions by copper metal,the standard cell potential at $25\,^{\circ}C$ is $0.46\,V$. What is the value of the standard Gibbs energy $(\Delta G^{\circ})$? $(F = 96500\,C\,mol^{-1})$

The $emf$ (in $V$) of a $Daniell$ cell containing $0.1 \ M \ ZnSO_4$ and $0.01 \ M \ CuSO_4$ solutions at their respective electrodes is $\left(E_{Cu^{2+} / Cu}^{\circ}=+0.34 \ V ; E_{Zn^{2+} / Zn}^{\circ}=-0.76 \ V\right)$

Find the $emf$ of the following cell reaction,given $E^0_{Cr^{3+}/Cr} = -0.74 \ V$ and $E^0_{Fe^{2+}/Fe} = -0.44 \ V$ at $25^{\circ} C$ for the cell: $Cr \ | \ Cr^{3+}(0.1 \ M) \ || \ Fe^{2+}(0.01 \ M) \ | \ Fe$. (in $V$)

What is the change in the reduction potential of a hydrogen electrode when the $pH$ of the initial solution changes from $0$ to $7$ by neutralization?