If a hydrogen electrode is dipped in two solutions of $pH = 3$ and $pH = 6$ and a salt bridge is connected,the e.m.f. of the resulting cell is ............ $V$.

What would be the electrode potential for the given half-cell reaction at $pH = 5$ (in $V$)?
$2 H_2O \rightarrow O_2 + 4 H^{\oplus} + 4 e^{-} ; E_{red}^{0} = 1.23 \ V$
$(R = 8.314 \ J \ mol^{-1} \ K^{-1} ; \text{Temperature} = 298 \ K ; \text{oxygen under std. atm. pressure of } 1 \ bar)$

Calculate the cell potential at $298 \ K$ for the following cells:
$(a)$ $Cd \mid Cd^{2+}(0.02 \ M) \parallel H^{+}(1 \ M) \mid H_{2(g)}(1 \ bar) \mid Pt$ $\left[ E_{Cd^{2+} \mid Cd}^0 = -0.40 \ V \right]$
$(b)$ $Al \mid Al^{3+}(0.25 \ M) \parallel Zn^{2+}(0.15 \ M) \mid Zn_{(s)}$ $\left[ E_{Al^{3+} \mid Al}^0 = -1.66 \ V, E_{Zn^{2+} \mid Zn}^0 = -0.76 \ V \right]$

For the cell reaction $Sn_{(s)} + 2Ag^+_{(aq)} \rightarrow Sn^{2+}_{(aq)} + 2Ag_{(s)}$,what happens when the cell voltage increases?

The Nernst equation is related to:

The standard cell potential for $Zn \mid Zn^{2+} \parallel Cu^{2+} \mid Cu$ is $1.10 \ V$. When the cell is completely discharged,$\log ([Zn^{2+}] / [Cu^{2+}])$ is closest to $.....$