The standard electrode potential for the following reaction is $+1.33 \ V$. What is the potential at $pH = 2.0$ for the reaction: $Cr_2O_7^{2-} (aq, 1 \ M) + 14H^{+} (aq) + 6e^{-} \rightarrow 2Cr^{3+} (aq, 1 \ M) + 7H_2O (l)$?

The electrode potential of the following half cell at $298 \ K$ is given by the cell reaction:
$X | X^{2+}(0.001 \ M) || Y^{2+}(0.01 \ M) | Y$
The cell potential is $....... \times 10^{-2} \ V$ (Nearest integer).
Given: $E^0_{X^{2+} | X} = -2.36 \ V$,$E^0_{Y^{2+} | Y} = +0.36 \ V$,$\frac{2.303 \ RT}{F} = 0.06 \ V$.

If the standard electrode potential for a cell is $2 \ V$ at $300 \ K,$ the equilibrium constant $(K)$ for the reaction $Zn_{(s)} + Cu^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Cu_{(s)}$ at $300 \ K$ is approximately $(R = 8 \ J \ K^{-1} \ mol^{-1}, F = 96000 \ C \ mol^{-1})$

Given the equilibrium constant $K_c$ of the reaction $Cu_{(s)} + 2Ag^{+}_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$ is $10 \times 10^{15}$,calculate the $E_{cell}^o$ of the reaction at $298 \ K$. [Given: $2.303 \ \frac{RT}{F} \text{ at } 298 \ K = 0.059 \ V$]

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 $Zn_{(s)} + Fe^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Fe_{(s)}$?

To find the standard potential of $M^{3+}/M$ electrode,the following cell is constituted:
$Pt | M | M^{3+} (0.001 \ mol \ L^{-1}) || Ag^{+} (0.01 \ mol \ L^{-1}) | Ag$
The $emf$ of the cell is found to be $0.421 \ V$ at $298 \ K$. The standard potential of the half-reaction $M^{3+} + 3e^{-} \to M$ at $298 \ K$ will be .............. $V$.
(Given $E^{o}_{Ag^{+}/Ag}$ at $298 \ K = 0.80 \ V$)