Consider the following cell reaction:
$2 Fe^{3+}_{(aq)} + 2 I^{-}_{(aq)} \rightleftharpoons 2 Fe^{2+}_{(aq)} + I_{2(s)}$
At $298 \ K$,the cell emf is $0.237 \ V$. The equilibrium constant for the reaction is $10^x$. The value of $x$ is:
$(F = 96500 \ C \ mol^{-1}; R = 8.3 \ J \ K^{-1} \ mol^{-1})$

What will be the electromotive force of the following cell $Fe|Fe^{+2}\,(0.2\, M)\,||\, Au^{+3}\,(0.02\, M)\,|Au$ (in $,V$)?
Given: $E^o_{Fe^{+2}/Fe} = -0.44\,V$ and $E^o_{Au^{+3}/Au} = +1.50\,V$.

At $298 \ K$ the equilibrium constant for the reaction $M_{(s)} + 2 Ag^{+}_{(aq)} \rightarrow M^{2+}_{(aq)} + 2 Ag_{(s)}$ is $10^{15}$. What is the $E_{cell}^{\ominus}$ (in $V$) for this reaction? $\left(\frac{2.303 RT}{F}\right) = 0.06 \ V$

The cell potential for $Zn | Zn^{2+}_{(aq)} || Sn^{x+}| Sn$ is $0.801 \ V$ at $298 \ K$. The reaction quotient for the above reaction is $10^{-2}$. The number of electrons involved in the given electrochemical cell reaction is .... (Given $E^{0}_{Zn^{2+}|Zn} = -0.763 \ V, E^{0}_{Sn^{x+}|Sn} = +0.008 \ V$ and $\frac{2.303 \ RT}{F} = 0.06 \ V$)

What pressure $(bar)$ of $H_2$ would be required to make the $emf$ of a hydrogen electrode zero in pure water at $25^{\circ} C$?

What is the potential of a cell containing two hydrogen electrodes,the negative one in contact with $10^{-8} \ M \ H^{+}$ and the positive one in contact with $0.025 \ M \ H^{+}$? (in $V$)