Calculate equilibrium constant for the equilibrium,

$2{\mathrm{MnO}}_4^{-}+6{\mathrm{H}}^{+}+5{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\rightleftharpoons 2{\mathrm{Mn}}^{2+}+8{\mathrm{H}}_2\mathrm{O}+10{\mathrm{CO}}_2$

Given that, ${\mathrm{E}}_{{\mathrm{MnO}}_4^{-}/ {\mathrm{Mn}}^{2+}}^0=1.51 \mathrm{V}$ and ${{\mathrm{E}}^0}_{{\mathrm{CO}}_2^{}/{\mathrm{C}}_2{\mathrm{O}}_4^{2-}}=-0.49 \mathrm{V}$. The value of equilibrium constant is ${10}^{\mathrm{x}}$, find the value of $\mathrm{x}$ upto the nearest integer.

In the galvanic cell $\mathrm{Cu}\left|{\mathrm{Cu}}^{2+}(1 \mathrm{M})\right| \left|{\mathrm{Ag}}^{+}(1 \mathrm{M})\right|\mathrm{Ag},$ the electrons will travel in the external circuit:

The standard reduction potentials at $298 \mathrm{K}$ for the following half-reactions are given against each.

$$\begin{gathered} {\mathrm{Zn}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Zn}; {\mathrm{E}}^0=-0.762 \mathrm{V} \\ {\mathrm{Cr}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{Cr}; {\mathrm{E}}^0=-0.74 \mathrm{V} \\ 2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2; {\mathrm{E}}^0=0.0 \mathrm{V} \\ {\mathrm{Fe}}^{3+}+{\mathrm{e}}^{-}\to {\mathrm{Fe}}^{2+}; {\mathrm{E}}^0=+0.77 \mathrm{V}. \end{gathered}$$

Which is the strongest reducing agent?

The standard reduction potentials, ${\mathrm{E}}^0$, for the half-reactions are as

$\begin{array}{l}\mathrm{Zn}\to {\mathrm{Zn}}^{2+}+2{\mathrm{e}}^{-}; {\mathrm{E}}^0=+0.76 \mathrm{V}\\ \mathrm{Fe}\to {\mathrm{Fe}}^{2+}+2{\mathrm{e}}^{-}; {\mathrm{E}}^0=+0.41 \mathrm{V}\end{array}$

the $\mathrm{emf}$ for the cell reaction,

${\mathrm{Fe}}^{2+}+\mathrm{Zn}\to {\mathrm{Zn}}^{2+}+\mathrm{Fe}$ is