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

Represent the cell in which the following reaction takes place:
$Mg_{(s)} + 2Ag^{+}(0.0001 \, M) \rightarrow Mg^{2+}(0.130 \, M) + 2Ag_{(s)}$
Calculate its $E_{cell}$ if $E^{\Theta}_{cell} = 3.17 \, V$.

If the $E^{\circ}_{cell}$ of an equilibrium reaction $A_{(s)} + 2B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + 2B_{(s)}$ at $298 \ K$ is $0.59 \ V$,the equilibrium constant $K_c$ is

Consider the following half-cell reaction: $Cr_2O_7^{2-}{_{\text{(aq)}}} + 14H^{+}{_{\text{(aq)}}} + 6e^{-} \rightarrow 2Cr^{3+}{_{\text{(aq)}}} + 7H_2O_{\text{(l)}}$. The reaction was conducted with the ratio $\frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}]} = 10^{-6}$. The $pH$ value at which the $EMF$ of the half-cell will become zero is $............$ (nearest integer value). [Given: standard half-cell reduction potential $E^{o}_{Cr_2O_7^{2-}, H^{+} / Cr^{3+}} = 1.33 \ V$,$\frac{2.303 RT}{F} = 0.059 \ V$]

In a cell,a copper electrode was used as a cathode. What is the electrode potential (in $V$) of the copper electrode dipped in $0.1 \ M \ Cu^{2+}$ solution at $298 \ K$?
$(E_{Cu^{2+}/Cu}^{\ominus} = 0.34 \ V; \frac{2.303 \ RT}{F} = 0.06 \ V)$

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