At $298 \ K$,if the $emf$ of the cell corresponding to the reaction,$Zn_{(s)} + 2H^+_{(aq)} \rightarrow Zn^{2+}(0.01 \ M) + H_{2(g)}(1 \ atm)$ is $0.28 \ V$,then the $pH$ of the solution at the hydrogen electrode is (Given: $\frac{2.303 \ RT}{F} = 0.06 \ V$,$E^o_{Zn^{2+}|Zn} = -0.76 \ V$)

For the cell,$Mn_{(s)}|Mn_{(aq)}^{2+}(0.4\,M)||Sn_{(aq)}^{2+}(0.04\,M)|Sn_{(s)}$,calculate the free energy change $(\Delta G)$ at $298\,K$ in $kJ$.
Given: $E_{Mn^{2+}|Mn}^o = -1.18\,V$; $E_{Sn^{2+}|Sn}^o = -0.14\,V$; $\frac{2.303\,RT}{F} = 0.06$

Calculate the cell potential for the following cell: ............... $V$
$Cr_{(s)} | Cr^{3+}_{(0.1 \, M)} || Fe^{2+}_{(0.01 \, M)} | Fe_{(s)}$
Given: $E^0_{Cr^{3+}|Cr} = -0.72 \, V$,$E^0_{Fe^{2+}|Fe} = -0.42 \, V$ (in $, V$)

For a general redox reaction: $aA + bB \xrightarrow{n e^-} cC + dD$. Derive the Nernst equation.

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

For a cell,$Cu_{(s)} \mid Cu^{2+}(0.001\,M) \mid\mid Ag^{+}(0.01\,M) \mid Ag_{(s)}$,the cell potential is found to be $0.43\,V$ at $298\,K$. The magnitude of standard electrode potential for $Cu^{2+}/Cu$ is $......... \times 10^{-2}\,V$. $[\text{Given}: E^{\Theta}_{Ag^{+}/Ag} = 0.80\,V \text{ and } \frac{2.303RT}{F} = 0.06\,V]$