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

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

In an electrochemical reaction of lead,at standard temperature,if $E^0_{(Pb^{2+}/Pb)} = m \ V$ and $E^0_{(Pb^{4+}/Pb)} = n \ V$,then the value of $E^0_{(Pb^{2+}/Pb^{4+})}$ is given by $m - x n$. The value of $x$ is $...........$. (Nearest integer)

For the given cell $Cu(s) | Cu^{2+}(C_1 \ M) || Cu^{2+}(C_2 \ M) | Cu(s)$,the change in Gibbs energy $(\Delta G)$ is negative,if:

What happens to the $emf$ of the cell $Zn_{(s)} | Zn^{+2} (1 \ M) || Ag^{+1} (1 \ M) | Ag_{(s)}$ if the concentration of $Ag^{+1}$ decreases to $0.1 \ M$?

Half cells $Zn|Zn^{2+}$ $(1 \, L, 0.1 \, M)$ and $Cu|Cu^{2+}$ $(1 \, L, x \, M)$ are connected to form a cell. Calculate the concentration of $Cu^{2+}$ in the solution when cell potential is $0.8 \, V$. $(E_{cell}^o = 1.1 \, V)$