The oxidation potential of a hydrogen electrode at $pH = 10$ and $P_{H_2} = 1 \, atm$ is ........... $V$.

$Emf$ of the following cell at $298 \,K$ in $V$ is $x \times 10^{-2}$. $Zn | Zn^{2+}(0.1 \,M) || Ag^{+}(0.01 \,M) | Ag$. The value of $x$ is .... . (Rounded off to the nearest integer) [Given: $E^{0}_{Zn^{2+}/Zn} = -0.76 \,V$; $E^{0}_{Ag^{+}/Ag} = +0.80 \,V$; $\frac{2.303 RT}{F} = 0.059$]

The $e.m.f.$ of the following galvanic cells are represented by $E_1, E_2, E_3$ and $E_4$. Which of the following statements is true?
$(i)$ $Zn|Zn^{2+} (1 \, M)||Cu^{2+} (1 \, M)|Cu$
$(ii)$ $Zn|Zn^{2+} (0.1 \, M)||Cu^{2+} (1 \, M)|Cu$
$(iii)$ $Zn|Zn^{2+} (1 \, M)||Cu^{2+} (0.1 \, M)|Cu$
$(iv)$ $Zn|Zn^{2+} (0.1 \, M)||Cu^{2+} (0.1 \, M)|Cu$

The $emf$ of a $Daniel$ cell at $298 \ K$ is ${E_1}$ for the cell reaction $Zn|ZnSO_4(0.01 \ M)||CuSO_4(1.0 \ M)|Cu$. When the concentration of $ZnSO_4$ is $1.0 \ M$ and that of $CuSO_4$ is $0.01 \ M$,the $emf$ changes to ${E_2}$. What is the relationship between ${E_1}$ and ${E_2}$?

Calculate the $emf$ of the cell: $Cr | Cr^{3+}(0.1 \ M) || Fe^{2+}(0.01 \ M) | Fe$. Given: $E^0_{Cr^{3+}/Cr} = -0.75 \ V$; $E^0_{Fe^{2+}/Fe} = -0.45 \ V$. Cell reaction: $2 \ Cr_{(s)} + 3 \ Fe^{2+}_{(aq)} \rightarrow 2 \ Cr^{3+}_{(aq)} + 3 \ Fe_{(s)}$.

Calculate the $E.M.F.$ of the following cell at $298 \ K$: $Zn_{(s)} | ZnSO_4(0.01 \ M) || CuSO_4(1.0 \ M) | Cu_{(s)}$ if $E^o_{cell} = 2.0 \ V$. (in $V$)