For the cell reaction,$3 Sn^{4+} + 2 Cr \longrightarrow 3 Sn^{2+} + 2 Cr^{3+}$,$E^{\circ}_{cell}$ is $0.89 \ V$. Then $\Delta G^{\circ}$ for the reaction is

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 $, V$)

Considering the cell $Cu | Cu^{2+} || Ag^{+} | Ag$,what happens to the $emf$ if the concentrations of both $Cu^{2+}$ and $Ag^{+}$ ions are increased by a factor of $10$?

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 ${E_2}$?

At what $pH$,given half cell $MnO_4^{-} (0.1 \ M) \mid Mn^{2+} (0.001 \ M)$ will have electrode potential of $1.282 \ V$? (Nearest Integer) Given $E_{MnO_4^{-} / Mn^{2+}}^{o} = 1.54 \ V, \frac{2.303 RT}{F} = 0.059 \ V$

The cell potential for the following reaction is $0.03305 \ V$ at $298 \ K$. Find the value of $x$ for the reaction: $Zn | Zn^{2+} (0.1 \ M) || Cd^{2+} (x \ M) | Cd$. (Given: $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$,$E^{\circ}_{Cd^{2+}/Cd} = -0.40 \ V$) (in $M$)